$y=\sqrt[3]{1+x}\Rightarrow {{y}^{3}}=1+x\xrightarrow{Moshtagh}3{{y}^{2}}{y}'=1\Rightarrow {y}'=\frac{1}{3{{y}^{2}}}$  $({y}'{)}'=(\frac{1}{3{{y}^{2}}}{)}'\Rightarrow \frac{-6y{y}'}{9{{y}^{4}}}=\frac{-2{y}'}{3{{y}^{3}}}=\frac{-2(\frac{1}{3{{y}^{2}}})}{3{{y}^{3}}}=\frac{-2}{9{{y}^{5}}}=k(\frac{1}{9{{y}^{5}}})\Rightarrow k=-2$