Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be defined as $f(x)=\begin{cases} e^{-1/x^2}sin{1/x}& \text{if}{\hspace .2 cm} x\neq 0\\ 0 & \text{if}{\hspace .2 cm} x=0 \end{cases}$ Then

$f'$ is continuous at x=0

when $x \neq 0, f'(x)=e^{-1/x^2}cos1/x(-1/x^2)+2/x^3sin1/xe^{-1/x^2}$

and $f'(0)=\lim_{x \to 0}f(x)-f(0)/x-0=\lim_{x \to 0}e^{-1/x^2}sin1/x/x$.

Now since $0\leq |e^{-1/x^2}sin1/x/x|\leq e^{-1/x^2}/x$

and $\lim_{x\to 0}e^{-1/x^2}/x=0$ we get

$lim_{x \to 0}e^{-1/x^2}sin1/x/x=0$.

So $f'(0)=0$.

Hence $f(x)=\begin{cases} \frac{2sin1/x-xcos1/x}{x^3e^{1/x^2}}& \text{if}{\hspace .2 cm} x\neq 0 0 & \text{if}{\hspace .2 cm} x=0 \end{cases}$

Since $|2sin1/x-xcos1/x|\leq 2+|x|$ and $\lim_{x \to 0}\frac{2+|x|}{x^3e^{1/x^2}}=0$

we get that $\lim_{x \to 0}f'(x)=0=f'(0)$.

So $f'$ is continuous at $x=0$.