Answer
$\frac{{dy}}{{dx}} = \frac{{2x - \cos y\cos x}}{{5 - \sin x\sin y}}$
Work Step by Step
$$\eqalign{
& \sin x\cos y = {x^2} - 5y \cr
& {\text{Differentiating both sides of the equation with respect to }}x \cr
& \underbrace {\frac{d}{{dx}}\left[ {\sin x\cos y} \right]}_{{\text{use product rule}}} = \frac{d}{{dx}}\left[ {{x^2}} \right] - \frac{d}{{dx}}\left[ {5y} \right] \cr
& \sin x\frac{d}{{dx}}\left[ {\cos y} \right] + \cos y\frac{d}{{dx}}\left[ {\sin x} \right] = \frac{d}{{dx}}\left[ {{x^2}} \right] - \frac{d}{{dx}}\left[ {5y} \right] \cr
& y{\text{ is a function of }}x,{\text{ using the chain rule we have}} \cr
& \sin x\left( { - \sin y} \right)\frac{{dy}}{{dx}} + \cos y\cos x = 2x - 5\frac{{dy}}{{dx}} \cr
& - \sin x\sin y\frac{{dy}}{{dx}} + \cos y\cos x = 2x - 5\frac{{dy}}{{dx}} \cr
& {\text{Collect the terms that contains }}\frac{{dy}}{{dx}} \cr
& 5\frac{{dy}}{{dx}} - \sin x\sin y\frac{{dy}}{{dx}} = 2x - \cos y\cos x \cr
& {\text{Solve this equation for }}\frac{{dy}}{{dx}} \cr
& \left( {5 - \sin x\sin y} \right)\frac{{dy}}{{dx}} = 2x - \cos y\cos x \cr
& \frac{{dy}}{{dx}} = \frac{{2x - \cos y\cos x}}{{5 - \sin x\sin y}} \cr} $$
