Answer
$$\frac{{dy}}{{dx}} = \frac{2}{{9\root 3 \of {{x^2}\left( {1 + {x^{1/3}}} \right)} }}$$
Work Step by Step
$$\eqalign{
& y = \root 3 \of {{{\left( {1 + {x^{1/3}}} \right)}^2}} \cr
& {\text{Write as}} \cr
& y = {\left( {{{\left( {1 + {x^{1/3}}} \right)}^2}} \right)^{1/3}} \cr
& y = {\left( {1 + {x^{1/3}}} \right)^{2/3}} \cr
& {\text{differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{2}{3}{\left( {1 + {x^{1/3}}} \right)^{ - 1/3}}\frac{d}{{dx}}\left[ {1 + {x^{1/3}}} \right] \cr
& {\text{Then}} \cr
& \frac{{dy}}{{dx}} = \frac{2}{3}{\left( {1 + {x^{1/3}}} \right)^{ - 1/3}}\left( {\frac{1}{3}{x^{ - 2/3}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{2}{{9{x^{2/3}}{{\left( {1 + {x^{1/3}}} \right)}^{1/3}}}} \cr
& \frac{{dy}}{{dx}} = \frac{2}{{9{{\left[ {{x^2}\left( {1 + {x^{1/3}}} \right)} \right]}^{1/3}}}} \cr
& \frac{{dy}}{{dx}} = \frac{2}{{9\root 3 \of {{x^2}\left( {1 + {x^{1/3}}} \right)} }} \cr} $$
