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It's also worth remembering that we don't strictly need a Taylor expansion in this case, because the purpose of a Taylor expansion is to linearize a function, which in this case we already know to be linear (as stated earlier in Einstein's paper, the transformation equations must be linear for space and time to be homogeneous).

So we know a priori that $\tau$ is of the form: $$ \tau\left(x',y,z,t\right) ~=~ Ax' + By + Cz + Dt + E \,,$$ and $$ \tau\left(0,0,0,t\right) ~=~ Dt+E \,,$$ where $D$ is $\frac{\nabla \tau}{\nabla t} .$ Likewise all the coefficients are simply the respective partial derivatives: $A=\frac{\nabla \tau}{\nabla x'}$, $B=\frac{\nabla \tau}{\nabla y}$, $C=\frac{\nabla \tau}{\nabla z}$, and $E$ is an additive constant.

So taking Einstein's $\frac{1}{2}\tau_0 + \tau_2 = \tau_1$ formula (your Eqn (1)), plugging in the coordinate arguments, and performing a little algebra, you can easily obtain his partial differential equation (your Eqn (3)): $$ \frac{\nabla \tau}{\nabla x'} + \frac{v}{c^2-v^2} \frac{\nabla \tau}{\nabla t} ~=~ 0 \,.$$ Note that during this algebraic process $x'$ simply cancels out anyway.

And the same process (which he omits in the paper), can be used to show algebraically that the partial derivatives with respect to $y$ and $z$ are zero, by considering a light ray that moves vertically (along $y$). It departs the origin at $\tau\left(0,0,0,t_0\right)$, reflects at $\tau\left(0,L,0,t_0+\frac{L}{\sqrt{c^2-v^2}}\right) ,$ and returns at $\tau\left(0,0,0,t_0+\frac{2L}{\sqrt{c^2-v^2}}\right) ,$ where $L$ is the vertical length in question. Same thing for $z$.

It's also worth remembering that we don't strictly need a Taylor expansion in this case, because the purpose of a Taylor expansion is to linearize a function, which in this case we already know to be linear (as stated earlier in Einstein's paper, the transformation equations must be linear for space and time to be homogeneous).

So we know a priori that $\tau$ is of the form: $$ \tau\left(x',y,z,t\right) ~=~ Ax' + By + Cz + Dt + E \,,$$ and $$ \tau\left(0,0,0,t\right) ~=~ Dt+E \,,$$ where $D$ is $\frac{\partial \tau}{\partial t} .$ Likewise all the coefficients are simply the respective partial derivatives: $A=\frac{\partial \tau}{\partial x'}$, $B=\frac{\partial \tau}{\partial y}$, $C=\frac{\partial \tau}{\partial z}$, and $E$ is an additive constant.

So taking Einstein's $\frac{1}{2}(\tau_0 + \tau_2) = \tau_1$ formula (your Eqn (1)), plugging in the coordinate arguments, and performing a little algebra, you can easily obtain his partial differential equation (your Eqn (3)): $$ \frac{\partial \tau}{\partial x'} + \frac{v}{c^2-v^2} \frac{\partial \tau}{\partial t} ~=~ 0 \,.$$ Note that during this algebraic process $x'$ simply cancels out anyway.

And the same process (which he omits in the paper), can be used to show algebraically that the partial derivatives with respect to $y$ and $z$ are zero, by considering a light ray that moves vertically (along $y$). It departs the origin at $\tau\left(0,0,0,t_0\right)$, reflects at $\tau\left(0,L,0,t_0+\frac{L}{\sqrt{c^2-v^2}}\right) ,$ and returns at $\tau\left(0,0,0,t_0+\frac{2L}{\sqrt{c^2-v^2}}\right) ,$ where $L$ is the vertical length in question. Same thing for $z$.

It's also worth remembering that we don't strictly need a Taylor expansion in this case, because the purpose of a Taylor expansion is to linearize a function, which in this case we already know to be linear (as stated earlier in Einstein's paper, the transformation equations must be linear for space and time to be homogeneous). So we know a priori that Tau is of the form: $\tau (x',y,z,t) = Ax' + By + Cz + Dt + E$, and $\tau(0,0,0,t) = Dt+E$, where $D$ is $\frac{\nabla \tau}{\nabla t}$. Likewise all the coefficients are simply the respective partial derivatives: $A=\frac{\nabla \tau}{\nabla x'}$, $B=\frac{\nabla \tau}{\nabla y}$, $C=\frac{\nabla \tau}{\nabla z}$, and $E$ is an additive constant.

So taking Einstein's $\frac{1}{2}\tau_0 + \tau_2 = \tau_1$ formula (your Eqn (1)), plugging in the coordinate arguments, and performing a little algebra, you can easily obtain his partial differential equation (your Eqn (3)): $\frac{\nabla \tau}{\nabla x'} + \frac{v}{c^2-v^2} \frac{\nabla \tau}{\nabla t} = 0$. Note that during this algebraic process $x'$ simply cancels out anyway.

And the same process (which he omits in the paper), can be used to show algebraically that the partial derivatives wrt y and z are zero, by considering a light ray that moves vertically (along y). It departs the origin at $\tau(0,0,0,t_0)$, reflects at $\tau(0,L,0,t_0+\frac{L}{\sqrt{c^2-v^2}})$, and returns at $\tau(0,0,0,t_0+\frac{2L}{\sqrt{c^2-v^2}})$, where $L$ is the vertical length in question. Same thing for $z$.

It's also worth remembering that we don't strictly need a Taylor expansion in this case, because the purpose of a Taylor expansion is to linearize a function, which in this case we already know to be linear (as stated earlier in Einstein's paper, the transformation equations must be linear for space and time to be homogeneous).

So we know a priori that $\tau$ is of the form: $$ \tau\left(x',y,z,t\right) ~=~ Ax' + By + Cz + Dt + E \,,$$ and $$ \tau\left(0,0,0,t\right) ~=~ Dt+E \,,$$ where $D$ is $\frac{\nabla \tau}{\nabla t} .$ Likewise all the coefficients are simply the respective partial derivatives: $A=\frac{\nabla \tau}{\nabla x'}$, $B=\frac{\nabla \tau}{\nabla y}$, $C=\frac{\nabla \tau}{\nabla z}$, and $E$ is an additive constant.

So taking Einstein's $\frac{1}{2}\tau_0 + \tau_2 = \tau_1$ formula (your Eqn (1)), plugging in the coordinate arguments, and performing a little algebra, you can easily obtain his partial differential equation (your Eqn (3)): $$ \frac{\nabla \tau}{\nabla x'} + \frac{v}{c^2-v^2} \frac{\nabla \tau}{\nabla t} ~=~ 0 \,.$$ Note that during this algebraic process $x'$ simply cancels out anyway.

And the same process (which he omits in the paper), can be used to show algebraically that the partial derivatives with respect to $y$ and $z$ are zero, by considering a light ray that moves vertically (along $y$). It departs the origin at $\tau\left(0,0,0,t_0\right)$, reflects at $\tau\left(0,L,0,t_0+\frac{L}{\sqrt{c^2-v^2}}\right) ,$ and returns at $\tau\left(0,0,0,t_0+\frac{2L}{\sqrt{c^2-v^2}}\right) ,$ where $L$ is the vertical length in question. Same thing for $z$.

It's also worth remembering that we don't strictly need a Taylor expansion in this case, because the purpose of a Taylor expansion is to linearize a function, which in this case we already know to be linear (as stated earlier in Einstein's paper, the transformation equations must be linear for space and time to be homogeneous). So we know a priori that Tau is of the form: Tau(x',y,z,t) = Ax' + By + Cz + Dt + E, and Tau(0,0,0,t) = Dt+E, where D is del(Tau)/del(t). Likewise all the coefficients are simply the respective partial derivatives: A=del(Tau)/del(x'), B=del(Tau)/del(y), C=del(Tau)/del(z), and E is an additive constant.
So taking Einstein's 1/2[Tau_0 + Tau_2] = Tau_1 formula (your Eqn (1)), plugging in the coordinate arguments, and performing a little algebra, you can easily obtain his partial differential equation (your Eqn (3)): del(Tau)/del(x') + (v/(c^2-v^2))*del(Tau)/del(t) = 0. Note that during this algebraic process x' simply cancels out anyway.

And the same process (which he omits in the paper), can be used to show algebraically that the partial derivatives wrt y and z are zero, by considering a light ray that moves vertically (along y). It departs the origin at Tau(0,0,0,t_0), reflects at Tau(0,L,0,t_0+L/sqrt(c^2-v^2), and returns at Tau(0,0,0,t_0+2L/sqrt(c^2-v^2)), where L is the vertical length in question. Same thing for z.

It's also worth remembering that we don't strictly need a Taylor expansion in this case, because the purpose of a Taylor expansion is to linearize a function, which in this case we already know to be linear (as stated earlier in Einstein's paper, the transformation equations must be linear for space and time to be homogeneous). So we know a priori that Tau is of the form: $\tau (x',y,z,t) = Ax' + By + Cz + Dt + E$, and $\tau(0,0,0,t) = Dt+E$, where $D$ is $\frac{\nabla \tau}{\nabla t}$. Likewise all the coefficients are simply the respective partial derivatives: $A=\frac{\nabla \tau}{\nabla x'}$, $B=\frac{\nabla \tau}{\nabla y}$, $C=\frac{\nabla \tau}{\nabla z}$, and $E$ is an additive constant.

So taking Einstein's $\frac{1}{2}\tau_0 + \tau_2 = \tau_1$ formula (your Eqn (1)), plugging in the coordinate arguments, and performing a little algebra, you can easily obtain his partial differential equation (your Eqn (3)): $\frac{\nabla \tau}{\nabla x'} + \frac{v}{c^2-v^2} \frac{\nabla \tau}{\nabla t} = 0$. Note that during this algebraic process $x'$ simply cancels out anyway.

And the same process (which he omits in the paper), can be used to show algebraically that the partial derivatives wrt y and z are zero, by considering a light ray that moves vertically (along y). It departs the origin at $\tau(0,0,0,t_0)$, reflects at $\tau(0,L,0,t_0+\frac{L}{\sqrt{c^2-v^2}})$, and returns at $\tau(0,0,0,t_0+\frac{2L}{\sqrt{c^2-v^2}})$, where $L$ is the vertical length in question. Same thing for $z$.