Answer
$$dy = 4{x^2}{\sec ^2}\left( {\frac{{{x^3}}}{3}} \right)dx$$
Work Step by Step
$$\eqalign{
& y = 4\tan \left( {\frac{{{x^3}}}{3}} \right) \cr
& {\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {4\tan \left( {\frac{{{x^3}}}{3}} \right)} \right] \cr
& \frac{{dy}}{{dx}} = 4{\sec ^2}\left( {\frac{{{x^3}}}{3}} \right)\frac{d}{{dx}}\left[ {\frac{{{x^3}}}{3}} \right] \cr
& \frac{{dy}}{{dx}} = 4{\sec ^2}\left( {\frac{{{x^3}}}{3}} \right)\left( {\frac{{3{x^2}}}{3}} \right) \cr
& \frac{{dy}}{{dx}} = 4{x^2}{\sec ^2}\left( {\frac{{{x^3}}}{3}} \right) \cr
& {\text{Write in differential form}} \cr
& dy = 4{x^2}{\sec ^2}\left( {\frac{{{x^3}}}{3}} \right)dx \cr} $$
