= Solution
At an <equilibrium point>, $x=-y^2$ and $x^2=a^2y$, so $y(y^3-a^2)=0$. With $d=|a|^{2/3}>0$, the two <equilibrium points> and their <Jacobian matrix> are
$$
\boxed{(0,0),\qquad(-d^2,d),}\qquad
J(x,y)=\begin{pmatrix}2x&-a^2\\1&2y\end{pmatrix}.
$$
At $(-d^2,d)$, $\det J=-3a^2<0$. The <trace-determinant stability criterion> therefore gives an unstable <saddle equilibrium>. At the origin, the <eigenvalues> are $\lambda=\pm i|a|$: its <linearization> is a <center equilibrium>. The <linearization stability theorem> does not classify this <nonhyperbolic equilibrium> of the nonlinear system.
\b[Nonlinear classification.] In fact the answer depends on $|a|$:
$$
\boxed{\begin{array}{c|c}
0<|a|<1&\text{unstable weak focus at the origin}\\
|a|=1&\text{stable center at the origin}\\
|a|>1&\text{asymptotically stable weak focus at the origin}.
\end{array}}
$$
Here is an explicit derivation, rather than an inference from the imaginary <eigenvalues>. When $a^2=1$, set $L=1-x+y$. Near the origin $L>0$, and the <first integral>
$$
H(x,y)=\frac{1+xy}{L}+\log L
$$
has
$$
H_x=\frac{x+y^2}{L^2},\qquad H_y=\frac{y-x^2}{L^2},\qquad
\frac{dH}{dt}=0,\qquad
H=1+\frac{x^2+y^2}{2}+O\bigl((x^2+y^2)^{3/2}\bigr).
$$
Its <Hessian matrix> at the origin is the identity. Small level curves are closed ovals on which the <vector field> is nonzero, proving the <center equilibrium> classification and <Lyapunov stability>.
For $a^2\ne1$, a higher-order <Lyapunov function> calculation settles the issue. Put $k=|a|$, $U=x$, $V=ky$, $\tau=kt$, $p=k^{-1}$ and $q=k^{-2}$. Then
$$
U_\tau=-V+pU^2,\qquad V_\tau=U+qV^2.
$$
Write $R^2=U^2+V^2$ and introduce the differential operators
$$
\mathcal L=-V\partial_U+U\partial_V,\qquad
\mathcal N=pU^2\partial_U+qV^2\partial_V.
$$
Successively cancel the terms of degrees three, four and five in the derivative of $R^2+H_3+H_4+H_5$, using these <homogeneous polynomials>:
$$
\begin{aligned}
H_3&=-2pU^2V-\tfrac43pV^3+\tfrac43qU^3+2qUV^2,\\
H_4&=2p^2U^2V^2-4pq(U^3V+UV^3)+(p^2-q^2)V^4,\\
H_5&=-\tfrac{4q}{15}(p^2+8q^2)U^5
+\tfrac{16q}{3}(p^2-q^2)U^3V^2
-\tfrac{4p}{3}(p^2-q^2)U^2V^3\\
&\quad+4q(p^2-q^2)UV^4+\tfrac{4p}{15}(11q^2-2p^2)V^5.
\end{aligned}
$$
Direct differentiation verifies $\mathcal LH_3=-\mathcal N R^2$, $\mathcal LH_4=-\mathcal NH_3$ and $\mathcal LH_5=-\mathcal NH_4$. The remaining degree-six term has circular average
$$
c=\frac1{2\pi}\int_0^{2\pi}(\mathcal NH_5)(\cos\theta,\sin\theta)\,d\theta
=\frac{pq(q^2-p^2)}4=\frac{1-k^2}{4k^7}.
$$
For example, the only nonzero averages of degree-six monomials needed here are $\langle U^6\rangle=\langle V^6\rangle=5/16$ and $\langle U^4V^2\rangle=\langle U^2V^4\rangle=1/16$. Since $\mathcal L=\partial_\theta$ in <polar coordinates>, the zero-average part can be cancelled by a degree-six <homogeneous polynomial> $H_6$ satisfying
$$
\mathcal LH_6=cR^6-\mathcal NH_5.
$$
Indeed its restriction to the unit circle is obtained by integrating a zero-average trigonometric polynomial; its Fourier modes have orders $2,4,6$, so multiplication by $R^6$ gives a polynomial. Consequently the positive <Lyapunov function> $W=R^2+H_3+H_4+H_5+H_6$ satisfies
$$
\frac{dW}{d\tau}=cR^6+O(R^7).
$$
For $k>1$ this is strictly negative near the origin, giving <asymptotic stability>. For $k<1$ it is strictly positive, and $W_\tau\geq C W^3$ in a sufficiently small neighbourhood; every nonzero starting point must eventually leave that neighbourhood. Finally $\theta_\tau=1+O(R)>0$, so the nearby trajectories rotate: these are <weak foci>, with the stability stated above. Merely calling the nonlinear origin a center for every $a\ne0$ would miss this distinction.
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