If $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3\mathrm{sgn}\left(\mathrm{x}\right)$ then

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3\mathrm{sgn}\left(\mathrm{x}\right)=\left\{\begin{array}{l}{\mathrm{x}}^3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em},\mathrm{x}>0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em},\mathrm{x}=0\\ -{\mathrm{x}}^3,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}<0\end{array}\right.$

clearly f(x) is continuous at x = 0

${\mathrm{f}}^1\left(\mathrm{x}\right)=\left\{\begin{array}{l}3{\mathrm{x}}^2,\mathrm{x}>0\\ 0,\mathrm{x}=0\\ -3{\mathrm{x}}^2,\mathrm{x}<0\end{array}\right.$

${\mathrm{f}}^1\left(0^{-}\right)={\mathrm{f}}^1\left(0^{+}\right)=0$    f(x) is derivable at x = 0