Answer
$$y = \frac{1}{3}{\tan ^3}x - \frac{1}{4}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = {\sec ^2}x{\tan ^2}x,{\text{ }}\left( {0, - \frac{1}{4}} \right) \cr
& {\text{Separate the variables}} \cr
& dy = {\sec ^2}x{\tan ^2}x \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {{{\tan }^2}x{{\sec }^2}x} dx \cr
& y = \frac{1}{3}{\tan ^3}x + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}\left( {0, - \frac{1}{4}} \right) \cr
& - \frac{1}{4} = \frac{1}{3}{\tan ^3}\left( 0 \right) + C \cr
& C = - \frac{1}{4} \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{1}{3}{\tan ^3}x - \frac{1}{4} \cr
& \cr
& {\text{Graph}} \cr} $$
