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Disclaimer: It might be a wrong question because $f(x)$ is differentiable at $x=0$

Given: $f(x)=x^{\frac{1}{3}}$

We have,

$(\mathrm{LHD}$ at $x=0)$

$\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$

$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h-0}$

$=\lim _{h \rightarrow 0} \frac{(0-h)^{\frac{1}{3}}-0^{\frac{1}{3}}}{-h}$

$=\lim _{h \rightarrow 0} \frac{(-h)^{\frac{1}{3}}}{-h}$

$=\lim _{h \rightarrow 0}(-h)^{\frac{-2}{3}}$

$=0$

(RHD at x = 0)

$\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$

$=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h-0}$

$=\lim _{h \rightarrow 0} \frac{(0+h)^{\frac{1}{3}}-0^{\frac{1}{3}}}{-h}$

$=\lim _{h \rightarrow 0} \frac{h^{\frac{1}{3}}}{h}$

$=\lim _{h \rightarrow 0} h^{\frac{-2}{3}}$

$=0$

LHD at (x = 0)= RHD at (x = 0)

Hence, $f(x)=x^{\frac{1}{3}}$ is differentiable at $x=0$