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Expand the function into separated terms: $f=g(x)+h(y)$, where $g=x^2-x^4$ and $h=y^8-y^4$. ThenPut $a=2^{-1/2}$ and $b=2^{-1/4}$. The nine critical points are all pairs from $x\in \{0,\pm a\}$ and $y\in \{0,\pm b\}$. The diagonal Hessian matrix has entries $2-12x^2$ and $56y^6-12y^2$.

At $(0,\pm b)$ both entries are positive, giving strict local minima of value $-1/4$. At $(\pm a,\pm b)$ they have opposite signs, giving four saddle points of value zero. At the three points on $y=0$ the Hessian is degenerate, so its sign alone does not classify them. Use the higher-order test for separated extrema: $h(y)=-y^4+O(y^8)$ is negative for small nonzero $y$. At $(0,0)$ the positive $x^2$ term and negative $y^4$ term give a saddle. At $(\pm a,0)$, $g-1/4=-(x^2-1/2{)}^2$, so every sufficiently small nonzero displacement decreases $f$ and these are strict local maxima.

For the level curves, the exact zero set consists of the two parabolas $x=\pm y^2$ and the oval $x^2+y^4=1$. They cross at the four nondegenerate saddle points. Small positive levels enclose each maximum; small negative levels enclose each minimum. The degenerate central saddle has quartic narrowing, locally $x^2-y^4=c$. Far along the $x$ axis the function is negative, while far along the $y$ axis it is positive. These signs and zero curves determine the connectivity shown in the sketch.

Let $f$ have continuous second partial derivatives near an interior critical point $x_{\ast }$ of an open subset of ${\mathbb{R}}^n$, with $\nabla f(x_{\ast })=0$. The Hessian matrix gives the following sufficient conditions:

A positive semidefinite or negative semidefinite Hessian matrix with a zero eigenvalue does not settle the classification. The higher-order test for separated extrema gives two quartic examples with exactly the same zero Hessian. These conditions apply to an interior critical point; a boundary optimum or a constrained optimization problem requires attention to the allowed directions. ^[ref]

The proof outline is the Taylor expansionFor a positive-definite matrix, the smallest eigenvalue supplies a positive bound proportional to $\Vert h{\Vert }^2$, which dominates the remainder for small enough $h$. Negate the function for the local maximum case. For opposite-sign eigenvalues, choose the two corresponding eigenvectors: the leading terms have opposite signs along their lines.

There are also necessary conditions. At an interior local minimum, the Hessian matrix is positive semidefinite; at an interior local maximum, it is negative semidefinite. Restriction to each straight line reduces this to the one-variable second derivative condition. Since the Laplacian is the trace of the Hessian, it is nonnegative at such a local minimum and nonpositive at such a local maximum. The sign of the Laplacian alone is insufficient: $x^2-2y^2$ has negative Laplacian but a saddle point of a scalar function at the origin.