Solution
= Solution
For $x\ne0$, $V'(x)=2x^{-3}e^{-1/x^2}$ never vanishes. At $x=0$, the smooth flat extension has $V'(0)=0$, and $V(x)>V(0)=0$ for every $x\ne0$. Hence \b[$x=0$ is the unique equilibrium and is stable].
= Solution
For $x\ne0$, $V'(x)=2x^{-3}e^{-1/x^2}$ never vanishes. At $x=0$, the smooth flat extension has $V'(0)=0$, and $V(x)>V(0)=0$ for every $x\ne0$. Hence \b[$x=0$ is the unique equilibrium and is stable].