Answer
$$y = - \frac{1}{{2\tan 2x - 3}}$$
Work Step by Step
$$\eqalign{
& y' = 4{y^2}{\sec ^2}2x\,,\,\,\,\,\,\,\,{\text{initial condition }}y\left( {\pi /8} \right) = 1 \cr
& {\text{write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = 4{y^2}{\sec ^2}2x \cr
& \cr
& {\text{separating the variables}} \cr
& dy = 4{y^2}{\sec ^2}2xdx \cr
& \frac{1}{{{y^2}}}dy = 4{\sec ^2}2xdx \cr
& \cr
& {\text{Integrate both sides of the equation}} \cr
& \int {\frac{1}{{{y^2}}}} dy = 2\int {{{\sec }^2}2x} \left( 2 \right)dx \cr
& - \frac{1}{y} = 2\tan 2x + C \cr
& \cr
& {\text{Use the initial condition }}y\left( {\pi /8} \right) = 1 \cr
& - \frac{1}{1} = 2\tan 2\left( {\frac{\pi }{8}} \right) + C \cr
& - 1 = 2\tan \left( {\frac{\pi }{4}} \right) + C \cr
& - 1 = 2\left( 1 \right) + C \cr
& C = - 3 \cr
& \cr
& {\text{Then}}{\text{, the particular solution is}} \cr
& - \frac{1}{y} = 2\tan 2x - 3 \cr
& {\text{solving for }}y \cr
& y = - \frac{1}{{2\tan 2x - 3}} \cr} $$
