= Solution
Expand the function into separated terms: $f=g(x)+h(y)$, where $g=x^2-x^4$ and $h=y^8-y^4$. Then
$$
f_x=2x(1-2x^2),\qquad f_y=4y^3(2y^4-1).
$$
Put $a=2^{-1/2}$ 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>.
\b[There are two maxima at $(\pm a,0)$, two minima at $(0,\pm b)$, and five <saddle points>: $(0,0)$ and $(\pm a,\pm b)$.]
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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ia/paper-2-critical-contours.png]
{title=Level curves, exact zero set and all nine classified critical points of the separated quartic-octic function}
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