= Solution
As $x\to0$ through nonzero values, $1/x^2\to+\infty$ and $e^{-1/x^2}\to0=f(0)$. Thus $f$ is <continuous> at zero.
For the <derivative> at zero use its definition, rather than substituting into a formula obtained away from zero:
$$
f'(0)=\lim_{h\to0}\frac{e^{-1/h^2}}h=0.
$$
Indeed, putting $t=1/|h|$ makes the <absolute value> $te^{-t^2}$, which tends to zero since <Gaussian decay dominates every power>. Therefore \b[$f$ is <continuous> and <differentiable> at zero, with $f'(0)=0$]. Away from zero, the <chain rule> gives $f'(x)=2x^{-3}e^{-1/x^2}$.
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