Answer
$\frac{{dy}}{{dx}} = \frac{{{e^y} + 2x\sin \left( {{x^2} + {y^2}} \right)}}{{ - x{e^y} - 2y\sin \left( {{x^2} + {y^2}} \right)}}$
Work Step by Step
$$\eqalign{
& \cos \left( {{x^2} + {y^2}} \right) = x{e^y} \cr
& {\text{Differentiating both sides of the equation with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\cos \left( {{x^2} + {y^2}} \right)} \right] = \underbrace {\frac{d}{{dx}}\left[ {x{e^y}} \right]}_{{\text{use product rule}}} \cr
& \frac{d}{{dx}}\left[ {\cos \left( {{x^2} + {y^2}} \right)} \right] = x\frac{d}{{dx}}\left[ {{e^y}} \right] + {e^y}\frac{d}{{dx}}\left[ x \right] \cr
& y{\text{ is a function of }}x,{\text{ using the chain rule we have}} \cr
& - \sin \left( {{x^2} + {y^2}} \right)\frac{d}{{dx}}\left[ {{x^2} + {y^2}} \right] = x{e^y}\frac{{dy}}{{dx}} + {e^y}\left( 1 \right) \cr
& - \sin \left( {{x^2} + {y^2}} \right)\left( {2x + 2y\frac{{dy}}{{dx}}} \right) = x{e^y}\frac{{dy}}{{dx}} + {e^y} \cr
& - 2x\sin \left( {{x^2} + {y^2}} \right) - 2y\sin \left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} = x{e^y}\frac{{dy}}{{dx}} + {e^y} \cr
& {\text{Collect the terms that contains }}\frac{{dy}}{{dx}} \cr
& - x{e^y}\frac{{dy}}{{dx}} - 2y\sin \left( {{x^2} + {y^2}} \right)\frac{{dy}}{{dx}} = {e^y} + 2x\sin \left( {{x^2} + {y^2}} \right) \cr
& {\text{Solve this equation for }}\frac{{dy}}{{dx}} \cr
& \left[ { - x{e^y} - 2y\sin \left( {{x^2} + {y^2}} \right)} \right]\frac{{dy}}{{dx}} = {e^y} + 2x\sin \left( {{x^2} + {y^2}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{{e^y} + 2x\sin \left( {{x^2} + {y^2}} \right)}}{{ - x{e^y} - 2y\sin \left( {{x^2} + {y^2}} \right)}} \cr} $$
