Answer
$$\frac{{dy}}{{dx}} = \frac{{ - 1 - y{{\sec }^2}\left( {xy} \right)}}{{x{{\sec }^2}\left( {xy} \right)}}$$
Work Step by Step
$$\eqalign{
& x + \tan \left( {xy} \right) = 0 \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {\tan \left( {xy} \right)} \right) = 0 \cr
& \cr
& {\text{Solve derivatives}}{\text{, use the chain rule for }}\frac{d}{{dx}}\left( {\tan \left( {xy} \right)} \right) \cr
& 1 + {\sec ^2}\left( {xy} \right)\frac{d}{{dx}}\left( {xy} \right) = 0 \cr
& {\text{Use the product rule}} \cr
& 1 + {\sec ^2}\left( {xy} \right)\left( {x\frac{d}{{dx}}\left( y \right) + y\frac{d}{{dx}}\left( x \right)} \right) = 0 \cr
& 1 + {\sec ^2}\left( {xy} \right)\left( {x\frac{{dy}}{{dx}} + y} \right) = 0 \cr
& \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& x{\sec ^2}\left( {xy} \right)\frac{{dy}}{{dx}} + y{\sec ^2}\left( {xy} \right) = - 1 \cr
& x{\sec ^2}\left( {xy} \right)\frac{{dy}}{{dx}} = - 1 - y{\sec ^2}\left( {xy} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{ - 1 - y{{\sec }^2}\left( {xy} \right)}}{{x{{\sec }^2}\left( {xy} \right)}} \cr} $$
