Thus, \[f(x)=sgn {{x}^{3}}=sgn x,\] which is neither continuous nor derivable at 0.
Note that \[{f}'({{0}^{+}})=\underset{h\to {{0}^{+}}}{\mathop{\text{lim}}}\,\,\frac{f(0+h)-f(0)}{h}\]\[=\underset{h\to {{0}^{+}}}{\mathop{\text{lim}}}\,\,\frac{1-0}{h}\to \infty \] and \[{f}'({{0}^{-}})=\underset{h\to {{0}^{-}}}{\mathop{\text{lim}}}\,\,\frac{f(0-h)-f(0)}{h}\]\[=\underset{h\to {{0}^{-}}}{\mathop{\text{lim}}}\,\,\frac{-1-0}{h}\to \infty \]. \[{f}'({{0}^{+}})\ne {f}'({{0}^{-}})\], f is not derivable at \[x=0\].