Let $h(u,v)=f(u+v,u-v)$ and $f_{xx}=f_{yy}$ for every $(x,y)\in\mathbb{R}^2$. In addition, $f\in{C^2}.$ Show that $f(x,y)=U(x+y)+V(x-y)$.
I think applying the Taylor theorem could be useful.
$$f(x,y)=f(x+h_1,y+h_2)-\left(\frac{\partial{f(x,y)}}{\partial{x}}h_1+\frac{\partial{f(x,y)}}{\partial{y}}h_2\right)-\frac 1 2\left(\frac{\partial^2{f(x,y)}}{\partial^2{x}}h_1^2+\frac{\partial^2{f(x,y)}}{\partial{x}\partial{y}}h_1h_2+\frac{\partial^2{f(x,y)}}{\partial^2{y}}h_2^2\right)-R(h_1,h_2)$$
On
This is a d'Alembert form solution for the hyperbolic PDE $$ f_{xx} - f_{yy} = 0 $$ One changes to variables $$ \xi = x - y \\ \eta = x + y $$ and uses the chain rule to get $$ \frac{\partial f}{\partial x} = \left( \frac{\partial f}{\partial \xi} \right) \frac{\partial \xi}{\partial x} + \left( \frac{\partial f}{\partial \eta} \right) \frac{\partial \eta}{\partial x} = \frac{\partial f}{\partial \xi} + \frac{\partial f}{\partial \eta} \\ \frac{\partial f}{\partial y} = \left( \frac{\partial f}{\partial \xi} \right) \frac{\partial \xi}{\partial y} + \left( \frac{\partial f}{\partial \eta} \right) \frac{\partial \eta}{\partial y} = -\frac{\partial f}{\partial \xi} + \frac{\partial f}{\partial \eta} $$ and $$ \left( \frac{\partial}{\partial x}\frac{\partial}{\partial x} \right) f = \left( \frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta} \right) \left( \frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta} \right) f = \left( \frac{\partial}{\partial \xi} \right)^2 f + 2 \left( \frac{\partial}{\partial \xi} \frac{\partial}{\partial \eta} \right) f + \left( \frac{\partial}{\partial \eta} \right)^2 f \iff \\ f_{xx} = f_{\xi\xi} + 2 f_{\xi\eta} + f_{\eta\eta} \\ \left( \frac{\partial}{\partial y}\frac{\partial}{\partial y} \right) f = \left( -\frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta} \right) \left( -\frac{\partial}{\partial \xi} + \frac{\partial}{\partial \eta} \right) f = \left( \frac{\partial}{\partial \xi} \right)^2 f - 2 \left( \frac{\partial}{\partial \xi} \frac{\partial}{\partial \eta} \right) f + \left( \frac{\partial}{\partial \eta} \right)^2 f \iff \\ f_{yy} = f_{\xi\xi} - 2 f_{\xi\eta} + f_{\eta\eta} $$ This gives the transformed PDE: $$ f_{\xi\eta} = 0 $$ Integration regarding $\eta$ gives $$ f_\xi = C(\xi) $$ where $C = C(\xi)$, as this is only constant regarding $\eta$, so it can still be dependent on $\xi$, thus $C(\xi)$ instead of just $C$.
Another integration, now regarding $\xi$, gives $$ f = \underbrace{\int C(\xi) d\xi}_{E(\xi)} + D(\eta) = E(\xi) + D(\eta) = E(x - y) + D(x + y) $$ where $E(\xi)$ is an antiderivative of $C(\xi)$ and $D$ is constant regarding $\xi$, so it can still be $D(\eta)$. If we rename the introduced functions, we get $$ f = V(x-y) + U(x+y) $$
You define $$ \xi=x+y\\ \eta=x-y $$ then the partial differential equation becomes $$ \frac{\partial^2f}{\partial \xi\partial \eta}=0 $$ The only way you can satisfy it is that it is a sum of 2 functions $$ U(\xi)+V(\eta) $$ Such combination is always identically zero by the application of a second mixed derivative.