$f\left(x\right)=\{\begin{array}{cc}{x}^2\left(\frac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}\right);& x\ne 0\\ 0,& x=0\end{array}.$   Then

At  $x=0,$
$$\begin{gathered} L.H.L=\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f(0-h) \\ =\underset{h\to 0}{\lim }{h}^2\left(\frac{e^{-1/h}-e^{1/h}}{e^{-2/h}+e^{1/h}}\right) \\ =\underset{h\to 0}{\lim }{h}^2\left(\frac{e^{-2/h}-1}{e^{-2/h}+1}\right) \\ =0\left(\frac{0-1}{0+1}\right)=0 \\ R.H.L.=\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f(0+h) \\ =\underset{h\to 0}{\lim }{h}^2\left(\frac{e^{1/h}-e^{-1/h}}{e^{1/h}+e^{-1/h}}\right) \\ =\underset{h\to 0}{\lim }{h}^2\left(\frac{1-e^{-2/h}}{1+e^{-2/h}}\right) \end{gathered}$$
$=0\left(\frac{1-0}{1+0}\right)=0$ and  $f\left(0\right)=0$
$\therefore L.H.L=R.H.L=f\left(0\right)$
Hence, $f\left(x\right)$  is continuous at  $x=0.$
Also,  $L.H.D=\underset{h\to 0}{\lim }\frac{f(0-h)-f\left(0\right)}{-h}$
$=\underset{h\to 0}{\lim }\frac{h^2\frac{e^{-1/h}-e^{1/h}}{e^{-1/h}+e^{1/h}}-0}{-h}$
$$\begin{gathered} =-\underset{h\to 0}{\lim }h\frac{e^{-2/h}-1}{e^{-2/h}+1} \\ =0 \end{gathered}$$

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

$=0$
Hence, $f\left(x\right)$  is differentiable at $x=0$  and  $f\text{'}\left(0\right)=0$