I'll use the method of transformations.
If [tex]f_X(x)[/tex] denotes the PDF of [tex]X[/tex], and [tex]y=g(x)=x^3 \iff x=g^{-1}(y) = y^{1/3}[/tex], we have
[tex]f_Y(y) = f_X\left(g^{-1}(y)\right) \left|\dfrac{dg^{-1}}{dy}\right|[/tex]
[tex]\dfrac{dg^{-1}}{dy} = \dfrac13 y^{-2/3}[/tex]
[tex]\implies f_Y(y) = f_X\left(y^{1/3}\right) \left|\dfrac13 y^{-2/3}\right| = \boxed{\begin{cases} \dfrac53 y^{2/3} & \text{if } 0 \le y \le 1 \\ 0 & \text{otherwise} \end{cases}}[/tex]
