 Solution

Put $F(y)=(y^2-4)\arctan y$. Its zeros, and hence the equilibrium points, are $-2,0,2$. For a perturbation $y=y_{\ast }+\eta$, linear stability analysis gives $\dot{\eta }=F^{\prime }(y_{\ast })\eta +O({\eta }^2)$, withAt zero, $F^{\prime }(0)=-4$, so sufficiently small perturbations decay and $y=0$ is asymptotically stable. At either outer equilibrium point, $F^{\prime }(\pm 2)=4\arctan 2>0$, so $y=-2$ 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.