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For $w=x^3{e}^{-x^2-y^2}$, the first derivatives are $w_x=x^2(3-2x^2)e^{-x^2-y^2}$ and $w_y=-2y{x}^3{e}^{-x^2-y^2}$. Both vanish at the origin. The local expansion isThere are no constant, linear or quadratic terms, so every second derivative vanishes at the origin and the Hessian determinant is zero. The second-order test therefore cannot classify the point.

The cubic term does determine the behaviour: along $y=0$, $w(x,0)$ has the sign of $x$. Every neighbourhood contains values above and below $w(0,0)=0$, so it is neither a maximum nor a minimum. It is a nonisolated stationary point with nearby values of both signs. Indeed the whole line $x=0$ is stationary, so this is not an isolated nondegenerate quadratic saddle.

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.