Answer
$$\frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{1 + {x^6}}}$$
Work Step by Step
$$\eqalign{
& y = {\tan ^{ - 1}}\left( {{x^3}} \right) \cr
& {\text{differentiate using the formula }} \cr
& \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}}{\text{ }}\left( {{\text{see page 467}}} \right) \cr
& \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {{x^3}} \right)}^2}}}\frac{d}{{dx}}\left( {{x^3}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 + {{\left( {{x^3}} \right)}^2}}}\left( {3{x^2}} \right) \cr
& {\text{simplifying}}{\text{, }} \cr
& \frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{1 + {x^{\left( 3 \right)\left( 2 \right)}}}} \cr
& \frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{1 + {x^6}}} \cr} $$