Examine the continuity and differentiability of the function $f\left(x\right) = \begin{cases} xe^{-\left(\frac{1}{\left|x\right|}+\frac{1}{x}\right)}, & x \neq 0 \\ 0, & x = 0 \end{cases}$
$\begin{aligned}
\left|x\right| = \begin{cases}
x, & x > 0 \\
-x, & x < 0
\end{cases}
\end{aligned}$
$\begin{aligned}
\therefore \; \dfrac{1}{\left|x\right|} + \dfrac{1}{x} = \begin{cases}
\dfrac{1}{x} + \dfrac{1}{x} = \dfrac{2}{x}, & x > 0 \\\\
\dfrac{-1}{x} + \dfrac{1}{x} = 0, & x < 0
\end{cases}
\end{aligned}$
$\begin{aligned}
\therefore \; f\left(x\right) = \begin{cases}
xe^{-\frac{2}{x}}, & x > 0 \\
xe^{-0} = x, & x < 0 \\
0, & x = 0
\end{cases}
\end{aligned}$
$\begin{aligned}
\text{Left Hand Derivative} = Lf'\left(0\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(0-h\right)-f\left(0\right)}{-h} \\
& = \lim\limits_{h \to 0} \; \dfrac{-h-0}{-h} \\
& = 1
\end{aligned}$
$\begin{aligned}
\text{Right Hand Derivative} = Rf'\left(0\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(0+h\right)-f\left(0\right)}{h} \\
& = \lim\limits_{h \to 0} \; \dfrac{he^{-\frac{2}{h}}-0}{h} \\
& = \lim\limits_{h \to 0} \; e^{-\frac{2}{h}} \\
& = 0
\end{aligned}$
$\because$ $Lf'\left(0\right) \neq Rf'\left(0\right)$, the function is not differentiable at $x=0$.
$\because$ $Lf'\left(0\right)$ and $Rf'\left(0\right)$ are finite, $f\left(x\right)$ is continuous at $x=0$.