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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.

Let $G=x^3{e}^{-x^2-y^2}$ and $a=\sqrt{3/2}$. Its partial derivatives areso both vanish at $(a,0)$ and $(-a,0)$. Evaluating the second derivatives at a point $(x,0)$ with $x^2=3/2$ givesTherefore the Hessian determinant is $18x^2{e}^{-3}=27e^{-3}>0$. Its two diagonal entries are negative at $x=a$ and positive at $x=-a$. HenceThese are also global extrema: for positive $x$, the factor $e^{-y^2}\le 1$ maximizes the positive value at $y=0$, and differentiating $x^3{e}^{-x^2}$ fixes its positive maximum at $a$. Oddness in $x$ supplies the corresponding global minimum.

At a stationary point, the linear terms vanish, so $\Delta f\simeq \frac{1}{2}(A{h}^2+2Bhk+C{k}^2)$, with $A=f_{xx}$, $B=f_{xy}$, $C=f_{yy}$. Let $D=AC-B^2$ be the Hessian determinant. For $A\ne 0$, completing the square givesThe second-derivative test therefore yieldsWhen $D<0$ the quadratic form is indefinite, including the case $A=0$, $B\ne 0$. When $D=0$, the quadratic approximation is degenerate and the test is inconclusive: higher-order terms may produce a minimum, maximum or saddle. For nondegenerate cases the quadratic term dominates the Taylor remainder sufficiently close to the point.

Write $h=x-x_0$, $k=y-y_0$, and let $a=f_{xx}$, $b=f_{xy}$ and $c=f_{yy}$ be evaluated at the critical point. Assuming $f$ is twice continuously differentiable nearby, the Taylor expansion isThe linear terms vanish because the gradient is zero. For $a>0$, completing the square givesConsequently the requested constant is the Hessian determinantThe displayed identity in the question is an identity for the quadratic part; for a general $f$, its right-hand side needs the remainder $o(h^2+k^2)$.

If $D>0$, the quadratic form is positive for every nonzero $(h,k)$. On the unit circle it has a positive minimum, so it dominates the remainder sufficiently near the critical point. Hence the point is a strict local minimum.

If $D<0$, along $k=0$ the quadratic term is $a{h}^2/2>0$, whereas along $h=-bk/a$ it is $D{k}^2/(2a)<0$. Along each line the remainder is of smaller order. Thus values both above and below $f(x_0,y_0)$ occur arbitrarily nearby, giving a saddle point.

The critical points satisfySubstitute $x=-y/2$ to obtain $y(3y-1/2)=0$. Thus the only critical points are $(0,0)$ and $(-1/12,1/6)$.

The Hessian matrix and Hessian determinant areAt $(0,0)$, $D=-1$, so this is a saddle point. At $(-1/12,1/6)$, $D=1$ and $f_{xx}=2>0$, so this is a strict local minimum. ThereforeThe value at the local minimum is $-1/432$; there is no local maximum.

For a twice continuously differentiable function, an interior local extremum requires $f_x=f_y=0$. At such a critical point, put $D=f_{xx}{f}_{yy}-f_{xy}^2$. The second-derivative test gives a strict local minimum if $D>0$ and $f_{xx}>0$, a strict local maximum if $D>0$ and $f_{xx}<0$, and a saddle point if $D<0$. When $D=0$, the test is inconclusive. More generally the Hessian matrix is necessarily positive semidefinite at a local minimum and negative semidefinite at a local maximum, but these non-strict signs alone are not sufficient.

For this polynomial,Subtracting the two stationarity equations gives $y(2x-y)=0$. If $y=0$, then $x=0$; if $y=2x$, the remaining equation gives $x=0$ or $x=3/2$. Thus the only critical points are $(0,0)$ and $(3/2,3)$.

At $(3/2,3)$, $\det H=-18<0$, so it is a saddle point. At $(0,0)$ the determinant is zero, but along $y=-x$ the value is $-x^3$, which takes both signs arbitrarily near zero. ThereforeIn particular a vanishing Hessian determinant does not exclude a saddle point.