Finding $f'(x)$ In the following problem I need to consider the function defined by $f(x)=e^{\frac{-1}{x^2}}$ if $x\ne 0$ and $0$ if $x=0$
I am trying to calculate $f'(x)$ for $x\ne0$. Then after calculating that use that to show that $f''(0)=0$
To calculate the first part I need to use the rule that,
$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ but how do I plug in the values I need then find my second part of my question?

Let $f$ be given by
$$f(x)=\begin{cases}e^{-1/x^2}&,x\ne 0\\\\0&,x=0\end{cases}$$

Then, for $x\ne0$, 
$$f'(x)=\frac{2e^{-1/x^2}}{x^3}$$
while for $x=0$ we see that
$$\begin{align}
f'(0)&=\lim_{x\to 0}\frac{e^{-1/x^2}-0}{x-0}\\\\
&=0
\end{align}$$

Finally, we have
$$\begin{align}
f''(0)&=\lim_{x\to 0}\frac{f'(x)-f'(0)}{x}\\\\
&=\lim_{x\to 0}\frac{\frac{2e^{-1/x^2}}{x^3}-0}{x-0}\\\\
&=0
\end{align}$$
