Answer
\[\frac{{dy}}{{dx}} = \frac{{y{e^{xy}}}}{{2 - x{e^{xy}}}}\]
Work Step by Step
\[\begin{gathered}
{e^{xy}} = 2y \hfill \\
\hfill \\
use\,\,the\,\,implicit\,\,differentiation \hfill \\
\hfill \\
\frac{d}{{dx}}\,\,\left[ {{e^{xy}}} \right] = \frac{d}{{dx}}\,\,\left[ {2y} \right] \hfill \\
\end{gathered} \]
\[differentiate\]
\[{e^{xy}}\,\,\left[ {xy' + y} \right] = 2y'\]
\[multiply\]
\[xy'\,{e^{xy}} + y{e^{xy}} = 2y'\]
\[collect\,\,like\,\,terms\]
\[xy'{e^{xy}} - 2y' = - y{e^{xy}}\]
\[solve\,\,for\,\,{y^,}\]
\[\begin{gathered}
y'\,\left( {x{e^{xy}} - 2} \right) = - y{e^{xy}} \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{y{e^{xy}}}}{{2 - x{e^{xy}}}} \hfill \\
\end{gathered} \]
