Answer
$$\frac{{dy}}{{dx}} = \frac{{y\cos x - \cos \left( {2x + 3y} \right) + 2x\sin \left( {2x + 3y} \right)}}{{ - 3x\sin \left( {2x + 3y} \right) - \sin x}}$$
Work Step by Step
$$\eqalign{
& x\cos \left( {2x + 3y} \right) = y\sin x \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {x\cos \left( {2x + 3y} \right)} \right) = \frac{d}{{dx}}\left( {y\sin x} \right) \cr
& \cr
& {\text{Use the product rule}} \cr
& x\frac{d}{{dx}}\left( {\cos \left( {2x + 3y} \right)} \right) + \cos \left( {2x + 3y} \right)\frac{d}{{dx}}\left( x \right) = y\frac{d}{{dx}}\left( {\sin x} \right) + \sin x\frac{d}{{dx}}\left( y \right) \cr
& \cr
& {\text{Find derivatives}} \cr
& - x\sin \left( {2x + 3y} \right)\frac{d}{{dx}}\left( {2x + 3y} \right) + \cos \left( {2x + 3y} \right) = y\cos x + \sin x\frac{{dy}}{{dx}} \cr
& - x\sin \left( {2x + 3y} \right)\frac{d}{{dx}}\left( {2x + 3y} \right) + \cos \left( {2x + 3y} \right) = y\cos x + \sin x\frac{{dy}}{{dx}} \cr
& - x\sin \left( {2x + 3y} \right)\left( {2 + 3\frac{{dy}}{{dx}}} \right) + \cos \left( {2x + 3y} \right) = y\cos x + \sin x\frac{{dy}}{{dx}} \cr
& \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& - 2x\sin \left( {2x + 3y} \right) - 3x\sin \left( {2x + 3y} \right)\frac{{dy}}{{dx}} + \cos \left( {2x + 3y} \right) = y\cos x + \sin x\frac{{dy}}{{dx}} \cr
& - 3x\sin \left( {2x + 3y} \right)\frac{{dy}}{{dx}} - \sin x\frac{{dy}}{{dx}} = y\cos x - \cos \left( {2x + 3y} \right) + 2x\sin \left( {2x + 3y} \right) \cr
& \left( { - 3x\sin \left( {2x + 3y} \right) - \sin x} \right)\frac{{dy}}{{dx}} = y\cos x - \cos \left( {2x + 3y} \right) + 2x\sin \left( {2x + 3y} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{y\cos x - \cos \left( {2x + 3y} \right) + 2x\sin \left( {2x + 3y} \right)}}{{ - 3x\sin \left( {2x + 3y} \right) - \sin x}} \cr} $$
