$z=z\left(x,y\right)$ is defined on region $D$, $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ both exists, and $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}$ for all $\left(x,y\right)\in D$.
If $z$ is differentiable we can get $z=f\left(x+y\right)$, i.e. z is only dependent on $x+y$.
But what if $z$ is not differentiable? Do there exist some cases when $z$ is not differentiable?
I guess that $z\left(x,y\right)$ must be differentiable if the conditions above are true, but anyway I cannot prove it. It's probable that I made a wrong guess.
Now it remains a problem: if $\frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}$ for all $\left(x,y\right)\in D$ and $z\left(x,x\right)\equiv 0$, then $z\left(x,y\right)\equiv 0$?
I think it would be true. But without given "$z$ is differentiable", it is hard to prove.
