 Solution

The first partial derivatives areBoth vanish at $(\lambda ,\lambda )$, so every positive diagonal point is a stationary point. The Hessian matrix there isIts eigenvectors $(1,1)$ and $(1,-1)$ have respective eigenvaluesThe zero eigenvalue reflects the entire line of nonisolated stationary points: $f(\lambda ,\lambda )=2$. In fact $f-2=(x-y{)}^2[1/(xy)-1/a^2]$, showing a non-strict local minimum when $\lambda <|a|$, a non-strict local maximum when $\lambda >|a|$, and values on both sides of two near $\lambda =|a|$. At that last point the whole Hessian matrix vanishes, so its signs alone cannot classify the point.