#### Answer

\[\frac{{dy}}{{dx}} = \frac{{1 - 3y\,{{\left( {xy + 1} \right)}^2}}}{{3x\,{{\left( {xy + 1} \right)}^2} + 2y}}\]

#### Work Step by Step

\[\begin{gathered}
\,{\left( {xy + 1} \right)^3} = x - {y^2} + 8 \hfill \\
\hfill \\
implicit\,\,differentiation \hfill \\
\hfill \\
\frac{d}{{dx}}\,\,\left[ {\,{{\left( {xy + 1} \right)}^3}} \right] = \frac{d}{{dx}}\,\,\left[ {x - {y^2} + 8} \right] \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\end{gathered} \]
\[3\,{\left( {xy + 1} \right)^2}\,\left( {xy' + y} \right) = 1 - 2yy'\]
\[multiply\]
\[3xy'\,{\left( {xy + 1} \right)^2} + 3y\,{\left( {xy + 1} \right)^2} = 1 - 2yy'\]
\[collect\,\,like\,\,terms\]
\[3xy'\,{\left( {xy + 1} \right)^2} + 2yy' = 1 - 3y\,{\left( {xy + 1} \right)^2}\]
\[factor\,\,{y^,}\]
\[y' = \,\,\left[ {3x\,{{\left( {xy + 1} \right)}^2} + 2y} \right] = 1 - 3y\,{\left( {xy + 1} \right)^2}\]
\[solve\,\,for\,\,{y^,}\]
\[\frac{{dy}}{{dx}} = \frac{{1 - 3y\,{{\left( {xy + 1} \right)}^2}}}{{3x\,{{\left( {xy + 1} \right)}^2} + 2y}}\]