#### Answer

\[y' = - \frac{1}{{2y - 2\cos y\sqrt {x + {y^2}} }}\]

#### Work Step by Step

\[find\,\,\frac{{dy}}{{dx}}\,\,\,\,using\,\,implicit\,\,\,differentitation.\]
\[\frac{1}{{2\sqrt {x + {y^2}} }}\, \cdot \,\,\,\,{\left( {x + {y^2}} \right)^\prime } = \cos y \cdot y'\]
\[then\]
\[\frac{1}{{2\sqrt {x + {y^2}} \, \cdot \,\left( {1 + 2yy'} \right)}} = y'\cos y\]
\[simplify\]
\[\frac{{2yy'}}{{2\sqrt {x + {y^2}} }} - y'\cos y = - \frac{1}{{2\sqrt {x + {y^2}} }}\]
\[factor\,\,y'\]
\[y'\,\left( {\frac{y}{{\sqrt {x + {y^2}} }} - \cos y} \right) = - \frac{1}{{2\sqrt {x + {y^2}} }}\]
\[solve\,\,for\,\,y'\]
\[\begin{gathered}
y' = - \frac{{\frac{1}{{2\sqrt {x + {y^2}} }}}}{{\frac{y}{{\sqrt {x + {y^2}} }} - \cos y}} \hfill \\
\hfill \\
y' = - \frac{1}{{2y - 2\cos y\sqrt {x + {y^2}} }} \hfill \\
\end{gathered} \]