= Solution
At a <fixed point>, the first equation gives $y=-x$. Substitution into the second gives $2x(x^2-1)=0$, so the <equilibrium points> are $(0,0),(1,-1),(-1,1)$. The <Jacobian matrix> is
$$
J(x,y)=\begin{pmatrix}1&1\\-1-4xy&1-2x^2\end{pmatrix}.
$$
At the origin its <eigenvalues> are $1\pm i$, so \b[the origin is an unstable spiral]. The local rotation is clockwise: on the positive $x$-axis the velocity points downward. At either nonzero <equilibrium point>,
$$
J=\begin{pmatrix}1&1\\3&-1\end{pmatrix},\qquad \det(J-\lambda I)=\lambda^2-4.
$$
Thus \b[both nonzero equilibria are saddles], with <eigenvalues> $-2,2$. The <stable manifold> is tangent to $(1,-3)$, and the <unstable manifold> to $(1,1)$, at each <saddle equilibrium>. Since all these <equilibrium points> are hyperbolic, <linear stability of a planar equilibrium> gives their actual local types.
For the <phase portrait>, the <nullclines> are $y=-x$ and $y=x/(1-2x^2)$, the latter defined away from $x=\pm1/\sqrt2$. The horizontal direction has the sign of $x+y$; the vertical direction has the sign of $(1-2x^2)y-x$. The system is invariant under $(x,y)\mapsto(-x,-y)$, so trajectories occur in centrally symmetric pairs. The arrows below show these directions; the highlighted <stable manifolds> enter the <saddle equilibria>, whereas the highlighted <unstable manifolds> leave them.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-2-planar-phase.png]
{title=Clockwise unstable spiral and two saddles with stable and unstable separatrices}
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