Answer
$$\frac{{dy}}{{dx}} = \frac{{ - 2x\sin \left( {{x^2}} \right) - {e^y}}}{{x{e^y}}}$$
Work Step by Step
$$\eqalign{
& \cos {x^2} = x{e^y} \cr
& {\text{By implicit differentiation}} \cr
& \frac{d}{{dx}}\left[ {\cos {x^2}} \right] = \frac{d}{{dx}}\left[ {x{e^y}} \right] \cr
& - \sin \left( {{x^2}} \right)\frac{d}{{dx}}\left[ {{x^2}} \right] = x\frac{d}{{dx}}\left[ {{e^y}} \right] + {e^y}\frac{d}{{dx}}\left[ x \right] \cr
& - \sin \left( {{x^2}} \right)\left( {2x} \right) = x{e^y}\frac{{dy}}{{dx}} + {e^y} \cr
& {\text{Solving for }}\frac{{dy}}{{dx}} \cr
& - 2x\sin \left( {{x^2}} \right) - {e^y} = x{e^y}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - 2x\sin \left( {{x^2}} \right) - {e^y}}}{{x{e^y}}} \cr} $$
