Consider the differential equation $ x'=f(x) $ where $ f(x)=\begin{cases} 0 & x = 0 \\[12pt] -x^3\sin\left( {\frac{1}{x}} \right) & x \ne 0 \end{cases} $
I have to study the equilibrium points. First, I've proved that $f(x) \in C^1(\mathbb{R})$. Then I've found that the equilibrium points are $x=0$ and $x_k = \frac{1}{k\pi}$, with $k \in \mathbb{Z} \setminus \{0\}$. The points $x_k$ can be easily classified (stable, asymptotically stable or unstable) because they are hyperbolic points ($f'(x_k) \ne 0$). What about $x=0$? I've read on Hale-Koçak that $x=0$ seems to be stable but not asymptotically stable.
I managed to prove that we cannot find a $\delta > 0$ s.t. $xf(x)<0$ for $0<\vert x\vert <\delta$: by a lemma on Hale-Koçak, this tells us the point is not asymptotically stable.
What about stability? I should prove that $ \exists \delta > 0, \, \vert x \vert < \delta \Rightarrow xf(x)\le 0 $
I don't manage to prove this. How would you do? Thanks for your help.