I asked a similar question some time back but am trying to work this from another angle.
In deriving the lagrangian of a free particle, we use the homogeneity of space to conclude that the lagrangian does not depend on its position vector $\vec{x}$. By homogeneity of space, I understand that if you displace the initial position of the particle by a vector $\vec{c}$, then all points on the trajectory of the particle are displaced by the same vector $\vec{c}$. Looking at the euler lagrange equation, considering a one degree of freedom case:
If $x_1(t)$ is a solution to the E-L equation corresponding to the initial condition $x(t_1)=X_1$, $$\frac{\partial L(x_1(t),\dot{x}_1(t),t)}{\partial x} - \frac{\mathrm{d} }{\mathrm{d} t} \frac{\partial L(x_1(t),\dot{x}_1(t),t)}{\partial \dot{x}}=0 \tag{1}$$ and $x_1(t) + c$ is also a solution to the E-L equation corresponding to the initial condition $x(t_1)=X_1 + c$, where $c$ is an infinitesimal displacement, $$\frac{\partial L(x_1(t)+c,\dot{x}_1(t),t)}{\partial x} - \frac{\mathrm{d} }{\mathrm{d} t} \frac{\partial L(x_1(t)+c,\dot{x}_1(t),t)}{\partial \dot{x}}=0 \tag{2}$$ then I'd like to prove that $$\frac{\partial L(x,\dot{x},t)}{\partial x}=0 \tag{3}$$
So far, I tried expanding $L(x_1(t)+c,\dot{x}_1(t),t)$ as a taylor series in powers of $c$. Since $c$ is very small, the linear terms in $c$ dominate. I can then reduce eqn. $(2)$ to: $$[L_{xx}-L_{xx\dot{x}}\,\dot{x} - L_{x\dot{x}\dot{x}}\,\ddot{x} - L_{x\dot{x}t}]\,c=0\tag{4}$$
I am not sure how I can proceed from here.
If I can prove that $\frac{\partial L(x,\dot{x},t)}{\partial x}=0$ for an infinitesimal displacement, I can imagine an infinity of such successive displacements $\vec{c}$, making $\frac{\partial L(x,\dot{x},t)}{\partial x}=0$ valid for finite displacements.