= Solution
The fixed-point conditions are $y(y^2-1)=0$ and $x=y^2$. They give $(0,0)$ and $(1,\pm1)$. The <Jacobian matrix> is
$$
J=\begin{pmatrix}0&3y^2-1\\1&-2y\end{pmatrix}.
$$
At $(1,1)$ its <eigenvalues> are $-1\pm\sqrt3$; at $(1,-1)$ they are $1\pm\sqrt3$. Both equilibria are <saddle equilibria>. At the origin the <eigenvalues> are $\pm i$. This linear calculation alone does not distinguish a nonlinear center from a weak focus.
To settle the origin and the global connections, set
$$
a_\pm=\frac{1\pm\sqrt3}{2},\qquad
F_\pm=x-1-a_\pm(y^2-1).
$$
Using $2a_\pm^2-2a_\pm-1=0$, direct differentiation gives $\dot F_\pm=-2a_\pm yF_\pm$. Hence each parabola $F_\pm=0$ is invariant and, away from them, the following is a <first integral>:
$$
\boxed{I=|F_+|^{-a_-}|F_-|^{a_+},\qquad\frac{dI}{dt}=0.}
$$
The two parabolas intersect precisely at the <saddle equilibria>. Their arcs between $y=-1$ and $y=1$ enclose the lens
$$
1+a_+(y^2-1)<x<1+a_-(y^2-1),\qquad |y|<1.
$$
Inside it $F_+>0$, $F_-<0$. The origin is a critical point of $\log I$ with negative-definite Hessian $-2\sqrt3$ times the identity. It is therefore a strict local maximum of this <first integral>, proving a nonlinear <center equilibrium>. The regular interior levels are closed periodic orbits, circulating anticlockwise. In fact the only critical point of $I$ inside the lens is the origin, so these levels fill the lens up to its boundary.
On $F_+=0$, $\dot y=(a_+-1)(y^2-1)$, so the interior arc runs from the upper <saddle equilibrium> to the lower. On $F_-=0$ it runs from the lower to the upper. Together they form a <heteroclinic cycle>. Their outer branches supply the remaining stable and unstable <separatrices>.
For the full exterior portrait, the two invariant parabolas partition the rest of the plane into four regions. Where both $F_\pm>0$, $\dot y>0$ everywhere; where both are negative, $\dot y<0$. In the region above both <saddle equilibria>, $y>1$ and $\dot x>0$; in the corresponding lower region, $y<-1$ and $\dot x<0$. These exterior trajectories are unbounded and cannot cross the <separatrices>. The <nullclines> are $y=0,\pm1$ for $\dot x=0$ and $x=y^2$ for $\dot y=0$.
\Image[/past-exam-of-the-flows-fluctuations-and-complexity-course-of-the-university-of-oxford/2016-reversible-flow-phase-portrait.png]
{title=Full cubic-quadratic phase portrait and a central zoom: an anticlockwise family of periodic orbits inside a heteroclinic lens, two saddles, and unbounded exterior trajectories}
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