If $f\left(x\right)=\left\{\begin{array}{l}x{e}^{-\left(\frac{1}{\left|x\right|}+\frac{1}{x}\right)}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em},x\ne 0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em},x=0\end{array}\right.$ then f(x) is

$f\left(0\right)=0;f\left(x\right)=x{e}^{-\left(\frac{1}{\left|x\right|}+\frac{1}{x}\right)}$

$$\begin{gathered} \text{ R.H.L. }\underset{h\to 0}{lim} (0+h)e^{-2/h}=\underset{h\to 0}{lim} \frac{h}{e^{2/h}}=0 \\ \text{ L.H.L. }\underset{h\to 0}{lim} (0-h)e^{-\left(\frac{1}{h}-\frac{1}{h}\right)}=0 \end{gathered}$$

therefore, f (x) is continuous. 

$\text{ R.H.D }=\underset{h\to 0}{lim} \frac{(0+h)e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}-0}{h}=0$

$\text{ L.H.D. }=\underset{h\to 0}{lim} \frac{(0-h)e^{-\left(\frac{1}{h}-\frac{1}{h}\right)}-0}{-h}=1$

therefore, $\text{ L.H.D. }\ne \text{ R.H.D. }$

$f\left(x\right)$ is not differentiable at x = 0.