let $f(x, y)$ function, and $f_{xx}, f_{xy}, f_{yx}, f_{yy}$ are continuous in $\mathbb{R}^2$, $f_{xx}\equiv f_{yy}$. $z(u, v)$ is defined as $z(u, v)=f(u+v, u-v)$, I need to show that $\frac{\partial z}{\partial u}$ is not dependent on $v$ and deduce that exist functions $h_1(t)$, $h_2(t)$ so that $z(u, v)=h_1(u) + h_2(v)$.
Because $f_{xx}, f_{xy}, f_{yx}, f_{yy}$ are continuous in $\mathbb{R}^2$ $\Longrightarrow$ $f_x, f_y$ are differentiable $\Longrightarrow$ $f$ is differentiable. let $x(u, v)=u+v$ and $y(u, v)=u-v$. So according to the chain rule $$\frac{\partial z}{\partial u} = f_x\cdot x_u + f_y\cdot y_u,\space\space \frac{\partial z}{\partial v} = f_x\cdot x_v + f_y\cdot y_v \Longrightarrow \frac{\partial z}{\partial u} = f_x + f_y,\space\space \frac{\partial z}{\partial v} = f_x - f_y$$ but I don't know how to progress from here, how can I show that $\frac{\partial z}{\partial u}$ doesn't depend on $v$?
I'm not sure that I follow the notation in your last comment, but I think you've got it:
$$\frac{\partial z}{\partial u} = \frac{\partial }{\partial u} f(u+v, u-v) = f_x(u+v,u-v) + f_y(u+v,u-v)$$
In the basis of $x$ and $y$, the direction of increasing $v$ is $\langle 1, -1 \rangle$, so $\frac{\partial }{\partial v} \left(\frac{\partial z(u,v)}{\partial u}\right) = \left(\nabla_{x,y} \left[ f_x(u+v,u-v) + f_y(u+v,u-v) \right] \right)\cdot \langle 1, -1 \rangle$.
$$\nabla_{x,y} \left[ f_x(u+v,u-v) + f_y(u+v,u-v) \right] = \langle f_{xx}(u+v,u-v) + f_{yx}(u+v,u-v), f_{xy}(u+v,u-v) + f_{yy}(u+v,u-v) \rangle$$
Since the second derivatives are continuous, by Clairaut's theorem $f_{xy} = f_{yx}$. Also, we are told that $f_{xx} = f_{yy}$
Therefore
$$\frac{\partial }{\partial v} \left(\frac{\partial z(u,v)}{\partial u}\right) = \left(f_{xx}(u+v,u-v) + f_{yx}(u+v,u-v)\right) - \left(f_{xy}(u+v,u-v) + f_{yy}(u+v,u-v)\right) = 0$$