= Solution
The simultaneous zeros are the origin and the four corners $(\pm1,\pm1)$. The <Jacobian matrix> is
$$
J=\begin{pmatrix}-2xy&1-x^2\\1-y^2&-2xy\end{pmatrix}.
$$
At the origin its <eigenvalues> are $1,-1$, with <eigenvectors> $(1,1),(1,-1)$, so the origin is a <saddle equilibrium> of <Poincaré index> $-1$. At a corner $J=-2xyI$. Thus \b[$(1,1)$ and $(-1,-1)$ are stable nodes; $(1,-1)$ and $(-1,1)$ are unstable nodes], all of index $+1$. Although the node <eigenvalues> coincide, the equilibria are hyperbolic and their stability follows from their strictly signed real <eigenvalues>.
The lines $x=\pm1$ and $y=\pm1$ are invariant and divide the plane into nine convex regions. A nonconstant periodic trajectory cannot cross any of these lines by uniqueness, and cannot be contained in one of them because one-dimensional autonomous flows have no nonconstant periodic solutions. It must therefore lie inside one region. The central region contains only the origin, whose index is $-1$; all other regions contain no equilibria. But a simple <periodic orbit> has index $+1$, equal to the sum of the enclosed equilibrium indices. Its interior stays in the same convex region, giving a contradiction in either case. \b[There are no <periodic orbits>.]
Away from the invariant lines, trajectory curves have the <first integral>
$$
\frac{1-x^2}{1-y^2}=C,
$$
since both logarithmic derivatives equal $-2xy$. The diagonal $y=x$ is the unstable <separatrix> of the origin, leading toward the two sinks; $y=-x$ is its stable <separatrix>, coming from the two sources. The diagram shows these separatrices, the invariant lines and the direction field.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ii/paper-2-phase-plane.png]
{title=Phase portrait with a central saddle, two sources, two sinks and invariant square boundaries}
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