= Solution
Put $F(y)=(y^2-4)\arctan y$. Its zeros, and hence the <equilibrium points>, are $-2,0,2$. For a perturbation $y=y_*+\eta$, <linear stability analysis> gives $\dot\eta=F'(y_*)\eta+O(\eta^2)$, with
$$
F'(y)=2y\arctan y+\frac{y^2-4}{1+y^2}.
$$
At zero, $F'(0)=-4$, so sufficiently small perturbations decay and \b[$y=0$ is asymptotically stable]. At either outer <equilibrium point>, $F'(\pm2)=4\arctan2>0$, so \b[$y=-2$ and $y=2$ are unstable]. The <phase line> confirms the result: $F$ has signs $-,+,-,+$ on the intervals cut out by $-2,0,2$, respectively. Flow points toward zero on both sides and away from both outer equilibria.
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