= Solution
For this <quartic potential with fourfold symmetry>, the first <derivatives> factor:
$$
f_x=2x(1-x^2-by^2),\qquad f_y=2y(1-y^2-bx^2).
$$
The <critical points> are the origin, the four axis points $(\pm1,0),(0,\pm1)$, and four points with both coordinates nonzero. At the latter, $x^2+by^2=y^2+bx^2=1$. Subtracting and using $b\ne1$ gives $x^2=y^2=1/(1+b)$. Thus the complete list is
$$
\boxed{(0,0),\quad(\pm1,0),(0,\pm1),\quad
\left(\pm\frac1{\sqrt{1+b}},\pm\frac1{\sqrt{1+b}}\right),}
$$
where the two signs in the final group are chosen separately.
The <Hessian> is
$$
D^2f=\begin{pmatrix}2-6x^2-2by^2&-4bxy\\-4bxy&2-6y^2-2bx^2\end{pmatrix}.
$$
At the origin it is $2I$, so there is a strict <local minimum> with value zero. At an axis point its <eigenvalues> are $-4$ and $2(1-b)$, giving a <saddle point of a scalar function> if $b<1$ and a strict <local maximum> if $b>1$. All four axis values are $1/2$.
At a diagonal point the <Hessian> has <eigenvalues>
$$
-4,\qquad-\frac{4(1-b)}{1+b}.
$$
Hence these four points are strict <local maxima> for $b<1$ and <saddle points of a scalar function> for $b>1$. Their value is $1/(1+b)$. None of these <critical points> is degenerate when $b\ne1$.
To describe the <level curves> precisely, use <polar coordinates> and write
$$
f=r^2-\frac12q(\theta)r^4,\qquad
q(\theta)=1+\frac{b-1}{2}\sin^2(2\theta)>0.
$$
For a level $f=c$, the possible radii satisfy
$$
\boxed{r^2=\frac{1\pm\sqrt{1-2q(\theta)c}}{q(\theta)},}
$$
retaining only real, nonnegative values. This formula, together with reflection in each axis and interchange of the coordinates, determines the <level curves>. The radial maximum in a direction is $1/(2q)$, attained at $r^2=1/q$. It also proves that the <local maxima> identified above are global: the largest directional value is $1/(1+b)$ on the diagonals if $b<1$, and $1/2$ on the axes if $b>1$.
For negative $c$ only the plus sign gives a positive radius, producing one outer closed <level curve>. At $c=0$ there is the isolated origin and the outer curve $r^2=2/q$. For $0<c<c_s$, where $c_s$ is the saddle value, both radial branches exist in every direction, giving an inner loop around the origin and an outer loop. At $c=c_s$ they meet at the four <saddle points of a scalar function>. Above the saddle value, four separate loops surround the four maxima; they shrink to those points at the maximum value. There are no <level curves> above it.
For \b[$0<b<1$], the saddle value is $c_s=1/2$ and the four high-level loops lie along the diagonals. For \b[$b>1$], $c_s=1/(1+b)$ and the four high-level loops lie along the coordinate axes. The sketches show representative values in both regimes; the highlighted <level curve> is the saddle level.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ia/paper-2-contours.png]
{title=Contours of the fourfold quartic potential, with diagonal maxima for b less than one and axis maxima for b greater than one}
Finally, at $b=1$ the expression depends only on radius:
$$
f=r^2-\frac12r^4=\frac12-\frac12(r^2-1)^2.
$$
Consequently $\boxed{\max f=1/2\text{, attained exactly on }x^2+y^2=1.}$ The maximum set is the whole unit circle, not merely four isolated points.
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