= Solution
Use the inverse <change of variables> $u=(x+y)/2$ and $v=(x-y)/2$. The <chain rule> gives the differential operators
$$
\partial_x=\tfrac12(\partial_u+\partial_v),\qquad \partial_y=\tfrac12(\partial_u-\partial_v).
$$
The <mixed partial derivatives> cancel, assuming the twice continuously differentiable solution appropriate here. Consequently $f_{xy}=(g_{uu}-g_{vv})/4$.
\b[The transformed equation is]
$$
\boxed{g_{uu}-g_{vv}=4.}
$$
The factors of $1/2$ belong to the inverse <Jacobian matrix>; using the forward transformation instead would give an incorrect factor.
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