Answer
$0$
Work Step by Step
$\lim\limits_{h \to 0} f'(0)=\lim\limits_{h \to 0 }\dfrac{f(0+h)-f(0)}{h})$
and $\lim\limits_{h \to 0 }\dfrac{1/h}{e^{1/h^2}}=\dfrac{\infty}{\infty}$
This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$
$\lim\limits_{h \to 0} \dfrac{\frac{-1}{h^2}}{e^{\frac{1}{h^2}}(\dfrac{-2}{h^2})}=\lim\limits_{h \to 0} (\dfrac{h}{2})( e^{-1/h^2})=0$
and $(0) \cdot (\infty)=0$
