= Solution
A <critical point of a dynamical system> satisfies $x=0$ and $y-2y^3=0$, so the critical points are
$$
(0,0),\qquad \left(0,\frac1{\sqrt2}\right),\qquad \left(0,-\frac1{\sqrt2}\right).
$$
The <Jacobian matrix> of the vector field is
$$
J(x,y)=\begin{pmatrix}0&1-6y^2\\-1&0\end{pmatrix}.
$$
At the origin its <eigenvalues> are $\pm i$. Moreover, $H=x^2+y^2+O(y^4)$ has a strict local minimum there, so the origin is a stable <center of a planar dynamical system>. At either other critical point,
$$
J=\begin{pmatrix}0&-2\\-1&0\end{pmatrix}
$$
has eigenvalues $\pm\sqrt2$, so both points are unstable <saddle points>. Their common value $H=1/4$ identifies the special <separatrices> shown below.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-2-phase-portrait.png]
{title=<Phase portrait> and trajectories of the dynamical system}
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