= Solution
For a <stationary point>, the two factors from part (a) must vanish. If $y=0$, the second equation gives $x(x^2-1)=0$; if $x=0$, the first gives $y(y^2-1)=0$. Thus the axis points are $(0,0),(\pm1,0),(0,\pm1)$.
If both coordinates are nonzero, solve $3x^2+y^2=1$ and $x^2+3y^2=1$, giving $x^2=y^2=1/4$. These four points complete the list of nine <stationary points>.
For classification, the <Hessian matrix> is
$$
H=\begin{pmatrix}6xy&3x^2+3y^2-1\\3x^2+3y^2-1&6xy\end{pmatrix}.
$$
At $(0,0)$ its <eigenvalues> are $1,-1$, and at the four unit-axis points they are $2,-2$. These five points are <saddle points of a scalar function>, all with value zero.
At $(1/2,1/2)$ and $(-1/2,-1/2)$ the <eigenvalues> are one and two, so both are strict <local minima> of value $-1/8$. At $(1/2,-1/2)$ and $(-1/2,1/2)$ they are minus one and minus two, so both are strict <local maxima> of value $1/8$. This follows from the <second-derivative test>. These extrema are local: the <polynomial> is unbounded above and below on the full <plane>.
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