Problem Statement:
Let $f:\mathbb{R}^2\to \mathbb{R}$ be a smooth function such that $f_{xx} = f_{yy}$ where we use the following notation $f_x = \frac{\partial f}{\partial x}.$ Then let $g(x,y) = f(x+y,x-y).$ What is the PDE satisfied by $g$?
My attempt:
We compute,
$$g_x = f_x + f_y \\
g_y = f_x - f_y
$$
and so,
$$g_{xy} = \frac{\partial}{\partial y}(f_{x} + f_y) = (f_{xx} - f_{xy}) + (f_{yx} - f_{yy})=0.$$
Now the solution to $g_{xy} = 0$ is $g(x,y) = A(x) + B(y) + K$ for some functions $A, B$ and constant $K$. Now as $g= f\circ \phi$ where $\phi(x,y) = (x+y, x-y)$ we get,
$$f= g\circ \phi^{-1} \implies f(x,y) = A\left(\frac{x+y}{2}\right)+B\left(\frac{x-y}{2}\right)+ K.$$ Is the application of chain rule in this computation correct?