Answer
\[\frac{{dy}}{{dx}} = \frac{{3{x^2}\,{{\left( {x - y} \right)}^2} + 2y}}{{2x}}\]
Work Step by Step
\[\begin{gathered}
{x^3} = \frac{{x + y}}{{x - y}} \hfill \\
\hfill \\
implicit\,\,differentiation \hfill \\
\hfill \\
\frac{d}{{dx}}\,\,\left[ {{x^3}} \right] = \frac{d}{{dx}}\,\,\left[ {\frac{{x + y}}{{x - y}}} \right] \hfill \\
\hfill \\
then \hfill \\
\hfill \\
3{x^2} = \frac{{\,\left( {x - y} \right)\,\left( {1 + y'} \right) - \,\left( {x + y} \right)\,\left( {1 - y'} \right)}}{{\,{{\left( {x - y} \right)}^2}}} \hfill \\
\hfill \\
mutliply \hfill \\
\hfill \\
3{x^2} = \frac{{x + xy' - y - yy' - x + xy' - y + yy'}}{{\,{{\left( {x - y} \right)}^2}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
3{x^2} = \frac{{2xy' - 2y}}{{\,{{\left( {x - y} \right)}^2}}} \hfill \\
\hfill \\
solve\,\,for\,\,{y^,} \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{{3{x^2}\,{{\left( {x - y} \right)}^2} + 2y}}{{2x}} \hfill \\
\end{gathered} \]
