Answer
$$y = \pm \frac{3}{{\sqrt {9\sin x - 1/2} }}$$
Work Step by Step
$$\eqalign{
& \left( {\sec x} \right)y'\left( x \right) = {y^3},{\text{ }}y\left( 0 \right) = 3 \cr
& {\text{Write }}y'\left( x \right){\text{ as }}\frac{{dy}}{{dx}} \cr
& \left( {\sec x} \right)\frac{{dy}}{{dx}} = {y^3} \cr
& {\text{Separate the variables}} \cr
& \frac{{dy}}{{{y^3}}} = \frac{1}{{\sec x}}dx \cr
& {y^{ - 3}}dy = \cos xdx \cr
& {\text{Integrate both sides}} \cr
& \int {{y^{ - 3}}} dy = \int {\cos x} dx \cr
& \frac{{{y^{ - 2}}}}{{ - 2}} = \sin x + C \cr
& - \frac{1}{{2{y^2}}} = \sin x + C \cr
& 2{y^2} = - \frac{1}{{\sin x + C}} \cr
& {\text{Use the initial condition }}y\left( 0 \right) = 3 \cr
& 2{\left( 3 \right)^2} = - \frac{1}{{\sin \left( 0 \right) + C}} \cr
& 18 = - \frac{1}{C} \cr
& C = - \frac{1}{{18}} \cr
& {\text{Substitute }}C{\text{ into }}2{y^2} = \frac{1}{{\sin x + C}} \cr
& 2{y^2} = \frac{1}{{\sin x - 1/18}} \cr
& {\text{Solve for }}y \cr
& 2{y^2} = \frac{{18}}{{18\sin x - 1}} \cr
& {y^2} = \frac{9}{{9\sin x - 1/2}} \cr
& y = \pm \frac{3}{{\sqrt {9\sin x - 1/2} }} \cr} $$
