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For this quartic potential with fourfold symmetry, the first derivatives factor:The critical points are the origin, the four axis points $(\pm 1,0),(0,\pm 1)$, and four points with both coordinates nonzero. At the latter, $x^2+b{y}^2=y^2+b{x}^2=1$. Subtracting and using $b\ne 1$ gives $x^2=y^2=1/(1+b)$. Thus the complete list iswhere the two signs in the final group are chosen separately.

The Hessian isAt 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 eigenvaluesHence 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 $b\ne 1$.

To describe the level curves precisely, use polar coordinates and writeFor a level $f=c$, the possible radii satisfyretaining 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 $0<b<1$, the saddle value is $c_s=1/2$ and the four high-level loops lie along the diagonals. For $b>1$, $c_s=1/(1+b)$ 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.

Finally, at $b=1$ the expression depends only on radius: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.