= Solution
At a <stationary point> of a twice continuously differentiable <function>, write the <Hessian matrix> as $H=\begin{pmatrix}A&B\\B&C\end{pmatrix}$ and its <determinant> as $D=AC-B^2$. The quadratic term in the <Taylor expansion> determines the local nature whenever $H$ is nonsingular.
If $D>0$, its two <eigenvalues> have the same sign. With $A>0$ the <quadratic form> is positive definite, giving a strict local minimum; with $A<0$ it is negative definite, giving a strict local maximum. If $D<0$, the <eigenvalues> have opposite signs and the point is a saddle, with nearby increasing and decreasing directions. Thus
$$
\boxed{D>0,\ A>0:\ \min;\qquad D>0,\ A<0:\ \max;\qquad D<0:\ \text{saddle}}.
$$
If $D=0$, the test is inconclusive. For example, $x^4+y^4$, $-x^4-y^4$ and $x^4-y^4$ all have zero <gradient> and zero <Hessian matrix> at the origin, yet give a minimum, a maximum and a saddle respectively. One must inspect higher-order terms or use another argument in such a case.
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