= Solution
{c}
The <equilibrium points> satisfy $y=\beta$ and $x(x^2-1)=0$, so they are
$$
\boxed{(-1,\beta),\qquad(0,\beta),\qquad(1,\beta).}
$$
The <Jacobian matrix> and its <eigenvalue> equation are
$$
J(x,y)=\begin{pmatrix}0&1\\3x^2-1&0\end{pmatrix},\qquad \lambda^2=3x^2-1.
$$
At $x=\pm1$ the <eigenvalues> are $\pm\sqrt2$, so both points are <saddle equilibria>. Their stable and unstable tangent directions have slopes $-\sqrt2$ and $\sqrt2$, respectively.
At $x=0$ the <eigenvalues> are $\pm i$, giving a centre for the <linearized equation>. Purely imaginary <eigenvalues> alone do not classify the nonlinear system. The <first integral> established in part (ii) proves that $(0,\beta)$ is in fact a <centre equilibrium>. It is clockwise: immediately to its right, with $0<x<1$ and $y=\beta$, the <vector field> points downwards. Thus the final nonlinear classification is
$$
\boxed{(-1,\beta),(1,\beta)\text{ are saddles};\quad(0,\beta)\text{ is a clockwise centre}.}
$$
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