Compute the derivative from the definition of $f(x)=e^{-1/x^2}$ for $x\neq 0.$ Is $f'(x)$ continuous? 
Compute the derivative from the definition of $f(x)=e^{-1/x^2}$ for $x\neq 0.$ Is $f'(x)$ continuous?

I tried using the definition of continuity and I wasn't sure how to show formally that $\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}=f'(x).$ Obviously, the derivative is $\dfrac{2}{x^3}e^{-1/x^2}.$
Here's my attempt.
So by the definition of the derivative, the derivative of $f(x)$ at a point $x$ is
$$\lim\limits_{h\to 0}\dfrac{e^{-1/(x+h)^2}-e^{-1/x^2}}{h}=e^{-1/x^2}\lim\limits_{h\to 0}\dfrac{e^{-1/(x+h)^2+1/x^2}-1}{h}\\
=e^{-1/x^2}\lim\limits_{h\to 0}\dfrac{e^{(2h+h^2)/(x^2(x+h)^2)}-1}{h},$$ but here I'm stuck. How can I get to the desired limit? Any help would be appreciated.
I know that $\lim\limits_{h\to 0} \dfrac{e^h-1}{h}=1.$ Also, we want to show that $$\lim\limits_{h\to 0}\dfrac{e^{(2hx+h^2)/(x^2(x+h)^2)}-1}{h}=\dfrac{2}{x^3}$$
I was thinking of using the Taylor series expansion for $e^x,$ but isn't there some other way? Thanks in advance!

Note that $$  \frac 1 {x^2} - \frac 1 {(x+h)^2} = \frac {h (2x + h)}{x^2 (x+h)^2} \xrightarrow {h \to 0} 0 \cdot \frac {2x}{x^4} =0,$$ so the difference of the two fractions is an infinitesimal. Now apply the limit you have learned: $$\frac {\mathrm e^h - 1}h \to 1 [h \to 0],$$
by substituting the infinitesimals to a simpler one according this limit expression. 
