Answer
$$y = {\sin ^{ - 1}}\left( {{e^{3\sin x + C}}} \right)$$
Work Step by Step
$$\eqalign{
& 3\tan y - \frac{{dy}}{{dx}}\sec x = 0 \cr
& \cr
& {\text{Add }}\frac{{dy}}{{dx}}\sec x{\text{ to both sides}} \cr
& 3\tan y = \frac{{dy}}{{dx}}\sec x \cr
& {\text{separating the variables}} \cr
& 3\tan ydx = \sec xdy \cr
& \frac{3}{{\sec x}}dx = \frac{1}{{\tan y}}dy \cr
& \cot ydy = 3\cos xdx \cr
& \cr
& {\text{Integrate both sides of the equation}} \cr
& \int {\cot y} dy = \int {3\cos x} dx \cr
& \int {\frac{{\cos y}}{{\sin y}}} dy = 3\int {\cos x} dx \cr
& \ln \left| {\sin y} \right| = 3\sin x + C \cr
& \cr
& {\text{Solve for }}y \cr
& \sin y = {e^{3\sin x + C}} \cr
& y = {\sin ^{ - 1}}\left( {{e^{3\sin x + C}}} \right) \cr} $$
