Solution
= Solution
The <chain rule> gives $u_{xx}=f''(x+\lambda y)$ and $u_{yy}=\lambda^2f''(x+\lambda y)$. The <partial differential equation> becomes
$$
(4-\lambda^2)f''(x+\lambda y)=0.
$$
For this to hold for every twice <differentiable> $f$, the coefficient must vanish:
$$
\boxed{\lambda=2\quad\text{or}\quad\lambda=-2}.
$$
For an affine $f$ alone any $\lambda$ would work, but the requirement is for arbitrary $f$.