Write the equation $f_{xx}-f_{yy} = 0$ in terms of the variables $u=x+y$, $v=x-y$ and calculate the general solution.
We define $g(u,v)=f(x,y)$, then, by the chain rule: $$f_x=g_u+g_v,f_y=g_u-g_v$$ Using the chain rule again $$f_{xx}=g_{uu}+g_{vu}+g_{uv}+g_{vv},f_{yy}=g_{uu}-g_{uv}-g_{vu}+g_{vv}$$
Therefore $$f_{xx}-f_{yy}=4g_{uv}$$ $$0=4g_{uv}$$
This means $g_u$ doesn't depend on $v$ and $g$...
I understand everything up until this point:
... has the form $\phi(u)+\psi(v)$ with $\phi$, $\psi$ functions of one variable. Therefore the general solution of $f_{xx}-f_{yy}=0$ is $$f(x,y)=\phi(x+y)+\psi(x-y)$$
I don't understand why $g$ has the form $\phi(u)+\psi(v)$ and how $g_u$ not depending on $v$ reflects on that solution
The function $g$ is a function of $(u,v)$, and its partial derivatives $g_u$, $g_v$ too. If $g_u$ is not dependent on $v$, then we can write $g_u = \phi'(u)$, for some function $\phi'$. Now, integrating $g_u$ with respect to $u$, we get $g = \phi(u) + \psi(v)$, where $\phi$ is an antiderivative of $\phi'$, and $\psi(v)$ is an "integration constant" (which is not constant here, but $v$-dependent). Finally, \begin{aligned} f(x,y) &= g(u,v) \\ &= \phi(u) + \psi(v) \\ &= \phi(x+y) + \psi(x-y) \, . \end{aligned}