Start from Euler (correctly written-- yours is not quite right) $$ \rho\left(\frac{\partial u}{\partial t}+ (u\cdot \nabla) u\right)= -\nabla (p+\rho g z) $$ and use $$ u_i\partial_i u_j= u_i \partial_j u_i + u_i(\partial_i u_j-\partial_j u_i) $$ in the form of the vector identity $$ (u\cdot \nabla) u=- u\times (\nabla\times u)+\nabla \left(\frac 12 |u|^2\right) $$ to write it as $$ \rho\left(\frac{\partial u}{\partial t}- u\times (\nabla\times u)\right)= -\nabla \left(p+\rho g z+\frac 12\rho|v^2|\right) $$ Now consider the steady flow in which $\partial u/\partial t=0$ and take a dot product with $u$ on both sides. You get $$ (v\cdot \nabla)\left(p+\rho g z+\frac 12\rho|v^2|\right)=0 $$ which is Bernoulli --- i.e the quanity inside the parentheses is constant along a streamline.

For compressible flow you need to write $$ \frac 1 \rho \nabla p= \nabla h $$ where $h$ is the specific enthalpy ($H=E+PV$ per unit mass) and then Bernoulli becomes $$ h+ g z+\frac 12|v^2|= constant. $$

Start from Euler (correctly written-- yours is not quite right) $$ \rho\left(\frac{\partial u}{\partial t}+ (u\cdot \nabla) u\right)= -\nabla (p+\rho g z) $$ and use $$ [(u\cdot \nabla) u]_j=(u_i\partial_i) u_j= u_i \partial_j u_i + u_i(\partial_i u_j-\partial_j u_i) $$ in the form of the vector identity $$ (u\cdot \nabla) u=- u\times (\nabla\times u)+\nabla \left(\frac 12 |u|^2\right) $$ to write it as $$ \rho\left(\frac{\partial u}{\partial t}- u\times (\nabla\times u)\right)= -\nabla \left(p+\rho g z+\frac 12\rho|u^2|\right) $$ Now consider the steady flow in which $\partial u/\partial t=0$ and take a dot product with $u$ on both sides. You get $$ (v\cdot \nabla)\left(p+\rho g z+\frac 12\rho|u^2|\right)=0 $$ which is Bernoulli --- i.e the quanity inside the parentheses is constant along a streamline.

For compressible flow you need to write $$ \frac 1 \rho \nabla p= \nabla h $$ where $h$ is the specific enthalpy ($H=E+PV$ per unit mass) and then Bernoulli becomes $$ h+ g z+\frac 12|v^2|= constant. $$