For function $f\left(x\right)$ at $x=0$
L.H.L. $=\underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(0-h\right)=\underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{2^{-\frac{1}{h}}-1}{2^{-\frac{1}{h}}+1}$$=\frac{0-1}{0+1}=-1$
R.H.L. $=\underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(0+h\right)=\underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{2^{\frac{1}{h}}-1}{2^{\frac{1}{h}}+1}=\underset{h\to 0+}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1-2^{-\frac{1}{h}}}{1+2^{-\frac{1}{h}}}$$=\frac{1-0}{1+0}=1$
Since, $\underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(0-h\right)\ne \underset{h\to 0^{+}}{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(0-h\right),$

therefore, $f\left(x\right)$ is not continuous and not differentiable at $x=0.$