Answer
$\frac{{dy}}{{dx}} = \frac{{2 - \cos x}}{{3 - \sin y}}$
Work Step by Step
$$\eqalign{
& \sin x + \cos y = 2x - 3y \cr
& {\text{Differentiating both sides of the equation with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\sin x} \right] + \frac{d}{{dx}}\left[ {\cos y} \right] = \frac{d}{{dx}}\left[ {2x} \right] - \frac{d}{{dx}}\left[ {3y} \right] \cr
& y{\text{ is a function of }}x,{\text{ using the chain rule we have}} \cr
& \cos x + \left( { - \sin y} \right)\frac{{dy}}{{dx}} = 2 - 3\frac{{dy}}{{dx}} \cr
& \cos x - \sin y\frac{{dy}}{{dx}} = 2 - 3\frac{{dy}}{{dx}} \cr
& {\text{Collect the terms that contains }}\frac{{dy}}{{dx}} \cr
& 3\frac{{dy}}{{dx}} - \sin y\frac{{dy}}{{dx}} = 2 - \cos x \cr
& {\text{Solve this equation for }}\frac{{dy}}{{dx}} \cr
& \left( {3 - \sin y} \right)\frac{{dy}}{{dx}} = 2 - \cos x \cr
& \frac{{dy}}{{dx}} = \frac{{2 - \cos x}}{{3 - \sin y}} \cr} $$
