Always solve such questions using the cumulative distribution approach. $$\begin{array}{rl}P(X\le x)& =\{\begin{array}{ll}0& x<0\\ \frac{1}{2}x& 0\le x\le 2\\ 1& 2<x\end{array}\\ P(Y\le y)& =P(X^3\le y)=P(X\le y^{\frac{1}{3}})\\ & =\{\begin{array}{ll}0& y<0\\ \frac{1}{2}{y}^{\frac{1}{3}}& 0\le y^{\frac{1}{3}}\le 2\\ 1& 2<y^{\frac{1}{3}}\end{array}\\ f_Y(y)& =\frac{dP(Y<=y)}{dy}(y)\\ & =\{\begin{array}{ll}0& y<0\\ \frac{1}{6}{y}^{\frac{-2}{3}}& 0\le y\le 8\\ 0& 8<y\end{array}\end{array}$$