#### Answer

\[y' = \frac{1}{{{e^y} + 2y\sin {y^2}}}\]

#### Work Step by Step

\[find\,\,\frac{{dy}}{{dx}}\,\,\,\,using\,\,implicit\,\,\,differentitation.\]
\[ - \sin \,{y^2} \cdot \,\,{\left( {{y^2}} \right)^\prime } + 1 = {e^y} \cdot y'\]
\[ - \sin \,\,{y^2} \cdot \,\,{\left( {{y^2}} \right)^\prime } + 1 = {e^y} \cdot y'\]
\[differentiate\]
\[y'{e^y} + 2yy'\sin {y^2} = 1\]
\[factor\,\,out\,\,y'\]
\[y'\,\left( {{e^y} + 2y\sin {y^2}} \right) = 1\]
\[solve\,\,for\,\,y'\]
\[y' = \frac{1}{{{e^y} + 2y\sin {y^2}}}\]