 Solution

The chain rule gives $u_{xx}=f^{\prime \prime }(x+\lambda y)$ and $u_{yy}={\lambda }^2{f}^{\prime \prime }(x+\lambda y)$. The partial differential equation becomesFor this to hold for every twice differentiable $f$, the coefficient must vanish:For an affine $f$ alone any $\lambda$ would work, but the requirement is for arbitrary $f$.