Answer
$$ - \tan 3x + C$$
Work Step by Step
$$\eqalign{
& - \int {3{{\sec }^2}3x} dx \cr
& {\text{set }}u = 3x{\text{ then }}\frac{{du}}{{dx}} = 3,\,\,\,\,\,\,\,\,\frac{{du}}{3} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& - \int {3{{\sec }^2}3x} dx = - \int {3{{\sec }^2}u} \left( {\frac{{du}}{3}} \right) \cr
& = - \int {{{\sec }^2}u} du \cr
& {\text{integrate by using the Basic Trigonometric integral }}\int {{{\sec }^2}x} dx = \tan x + C \cr
& = - \tan u + C \cr
& {\text{write in terms of }}x,{\text{ replacing }}3x{\text{ for }}u{\text{ gives}} \cr
& = - \tan 3x + C \cr} $$