Checking differentiability of $e^\frac{-1}{x}$ 
Let $f:\mathbb R\to\mathbb R$ is defined by
  $$f(x)=
\begin{cases}
e^\frac{-1}{x}&&x>0\\
0&&x\le0
\end{cases}$$

How do I check differentiability of $f(x)$ at $x=0$?
I have tried to use first principle but cannot proceed.

Use L'Hôpital's rule repeatedly as you approach $x = 0$ from the right:
\begin{align*}
\lim\limits_{x \to 0^+}f'(x) &= \lim\limits_{x \to 0^+}\frac{e^{-\frac{1}{x}}}{x^2} \\
&= \lim\limits_{x \to 0^+}\frac{\frac{1}{x^2}}{e^{\frac{1}{x}}} \quad\text{limit of the form $\frac{\infty}{\infty}$} \\
&= \lim\limits_{x \to 0^+}\frac{-\frac{2}{x^3}}{e^{\frac{1}{x}} \cdot \big(-\frac{1}{x^2}\big)} \\
&= \lim\limits_{x \to 0^+}\frac{\frac{2}{x}}{e^{\frac{1}{x}}} \quad\text{limit of the form $\frac{\infty}{\infty}$} \\
&= \lim\limits_{x \to 0^+}\frac{-\frac{2}{x^2}}{e^{\frac{1}{x}} \cdot \big(-\frac{1}{x^2}\big)} \\
&= \lim\limits_{x \to 0^+}\frac{2}{e^{\frac{1}{x}}} \\
&= \lim\limits_{x \to 0^+}2e^{-\frac{1}{x}} = 0 \\
\end{align*} Of course, approaching $x = 0$ from the left, $f(x)$ is just the constant function $0$; so $$\lim\limits_{x \to 0^-}f'(x) = \lim\limits_{x \to 0^-} 0 = 0$$ Thus, $f(x)$ is differentiable at $x = 0$ and the value of its derivative there is $0$.
