= Solution
Set both first <partial derivatives> equal to zero:
$$
f_x=4x^3-2x=2x(2x^2-1)=0,\qquad f_y=-4y^3+8y=-4y(y^2-2)=0.
$$
Thus $x=0,\pm1/\sqrt2$ and $y=0,\pm\sqrt2$, independently, giving exactly nine <stationary points>. The <Hessian matrix> is diagonal everywhere:
$$
H_f=\begin{pmatrix}12x^2-2&0\\0&8-12y^2\end{pmatrix}.
$$
Its diagonal entries are its <eigenvalues>, so the <second-derivative test> immediately classifies the points:
* At $(0,\pm\sqrt2)$, the <Hessian matrix> has <eigenvalues> $-2,-16$. These two points are strict <local maxima>, each with $f=4$.
* At $(\pm1/\sqrt2,0)$, the <Hessian matrix> has <eigenvalues> $4,8$. These two points are strict <local minima>, each with $f=-1/4$.
* At $(0,0)$, the <Hessian matrix> has <eigenvalues> $-2,8$, so it is a <saddle point of a scalar function> with $f=0$.
* At $(\pm1/\sqrt2,\pm\sqrt2)$, with all four sign combinations, the <Hessian matrix> has <eigenvalues> $4,-16$. These four points are <saddle points of a scalar function>, each with $f=15/4$.
In particular,
$$
\boxed{\text{2 local maxima, 2 local minima, and 5 saddle points.}}
$$
All nine <stationary points> are nondegenerate, so no higher-order test is needed. Neither type of <local extremum> is global: $f(x,0)=x^4-x^2\to+\infty$ as $|x|\to\infty$, whereas $f(0,y)=-y^4+4y^2\to-\infty$ as $|y|\to\infty$.
To organize the <contour plot>, complete the squares:
$$
f(x,y)=\frac{15}{4}+\left(x^2-\frac12\right)^2-(y^2-2)^2.
$$
The <level set> through all four outer <saddle points of a scalar function> factors exactly:
$$
f-\frac{15}{4}=\left(x^2-y^2+\frac32\right)\left(x^2+y^2-\frac52\right).
$$
It is therefore the union of the <circle> $x^2+y^2=5/2$ and the <hyperbola> $y^2-x^2=3/2$, which cross at those four <saddle points of a scalar function>. Near the origin, $f=-x^2+4y^2+O((x^2+y^2)^2)$, so the zero <contours> have tangents $y=\pm x/2$. These two exact structural features, together with the nested <contours> around the four <local extrema>, determine the requested sketch. They are an example of a <separable quartic landscape with nine stationary points>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-2-q12-quartic-contours.png]
{title=All nine stationary points and contours of the separable quartic}
{description=Original contour plot of $x^4-y^4-x^2+4y^2$. The two maxima, two minima and five saddles are marked. The exact level $15/4$ is highlighted as a circle and a hyperbola intersecting at the four outer saddles; the zero contour crosses at the origin.}
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