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The gradient factors asIf a coordinate is zero, the solutions are $(0,0)$, $(\pm 1,0)$ and $(0,\pm 1)$. With both coordinates nonzero, subtracting the two bracket equations gives $x^2=y^2$, hence $x^2=y^2=1/4$. These give four more stationary points.

The Hessian matrix has equal diagonal entries $6xy$ and off-diagonal entry $3(x^2+y^2)-1$. At the origin its eigenvalues are $\pm 1$, and at the four nonzero axis points they are $\pm 2$, so all five are saddles. At the two same-sign interior points the eigenvalues are $1,2$, while at the opposite-sign points they are $-1,-2$. ThereforeThe extrema are not global, since the quartic takes arbitrarily large positive and negative values along suitable rays.

For the contour sketch, the zero level is exactly the two axes together with the unit circle. Inside the circle, $f$ is negative in the first and third quadrants and positive in the other two; outside, those signs reverse. Small negative levels contain closed loops around the two minima as well as unbounded branches in the second and fourth quadrants. Small positive levels similarly contain two loops around the maxima and unbounded branches in the first and third quadrants. The closed loops shrink to the extrema at levels $\mp 1/8$, and disappear beyond those levels. This is the unit-scale quartic surface with axes and a circle as its zero set.