Answer
$$\frac{1}{2}\sin {y^2} = \tan x + C$$
Work Step by Step
$$\eqalign{
& yy' = \sec {y^2}{\sec ^2}x \cr
& y\frac{{dy}}{{dx}} = \sec {y^2}{\sec ^2}x \cr
& {\text{Separating variables gives}} \cr
& \frac{y}{{\sec {y^2}}}dy = {\sec ^2}xdx \cr
& y\cos {y^2}dy = {\sec ^2}xdx \cr
& {\text{Integrating, we get}} \cr
& \int {y\cos {y^2}} dy = \int {{{\sec }^2}x} dx \cr
& \frac{1}{2}\sin {y^2} = \tan x + C \cr} $$
