= Separable quartic landscape with nine stationary points
{title2=$F(x,y)=x^4-a x^2-y^4+b y^2,\quad a,b>0$}
The <gradient> of this <polynomial> vanishes exactly when $x=0,\pm\sqrt{a/2}$ and $y=0,\pm\sqrt{b/2}$, giving nine <stationary points>. Its diagonal <Hessian matrix> has entries $12x^2-2a$ and $2b-12y^2$. The <second-derivative test> therefore gives two strict <local maxima> at $(0,\pm\sqrt{b/2})$, with value $b^2/4$, and two strict <local minima> at $(\pm\sqrt{a/2},0)$, with value $-a^2/4$. The origin and the four points $(\pm\sqrt{a/2},\pm\sqrt{b/2})$ are <saddle points of a scalar function>, with values zero and $(b^2-a^2)/4$, respectively. Every <stationary point> is nondegenerate.
Completing the squares organizes the <contour plot>:
$$
F=\frac{b^2-a^2}{4}+\left(x^2-\frac a2\right)^2-\left(y^2-\frac b2\right)^2.
$$
The <level set> through the four outer <saddle points of a scalar function> factors as
$$
F-\frac{b^2-a^2}{4}=\left(x^2-y^2+\frac{b-a}{2}\right)\left(x^2+y^2-\frac{a+b}{2}\right).
$$
It is a <circle> together with a <hyperbola> when $a\ne b$; for $a=b$, the latter factor's complementary zero set is the pair of lines $y=\pm x$. Neither the <local maxima> nor the <local minima> are global, since the restrictions to the two coordinate axes are unbounded in opposite directions. The case $a=1$, $b=4$ gives four outer saddle values $15/4$ and an exact circle-hyperbola contour crossing at all four outer <saddle points of a scalar function>.
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