= Solution
{c}
One suitable <autonomous planar system> is
$$
\boxed{\dot x=y^2-1,\qquad \dot y=x^2-1.}
$$
Its <equilibrium points> are exactly the four combinations $(x,y)=(\pm1,\pm1)$. Its <Jacobian matrix> is
$$
J(x,y)=\begin{pmatrix}0&2y\\2x&0\end{pmatrix},\qquad \lambda^2=4xy.
$$
Thus $(-1,-1)$ and $(1,1)$ have <eigenvalues> $\pm2$ and are <saddle equilibria>. At the other two points the <eigenvalues> are $\pm2i$. To establish the nonlinear <centre equilibrium> classifications, use the <first integral>
$$
H(x,y)=\frac{x^3}{3}-x-\frac{y^3}{3}+y,\qquad
\dot H=(x^2-1)(y^2-1)-(y^2-1)(x^2-1)=0.
$$
Its <Hessian matrix> is $\operatorname{diag}(2x,-2y)$, a <negative-definite matrix> at $(-1,1)$ and a <positive-definite matrix> at $(1,-1)$. Thus each is a strict <local extremum>, and nearby <regular level sets> are <closed curves> supporting <periodic orbits>, by the <planar first integral criterion for a centre>.
Immediately to the right of $(-1,1)$, the <vector field> points down because $x^2-1<0$, so that centre is clockwise. Immediately to the right of $(1,-1)$ it points up because $x^2-1>0$, so that centre is counterclockwise. This verifies both the nonlinear classifications and the prescribed orientations, with exactly four <equilibrium points>.
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