#### Answer

\[{y^,} = \frac{{15{x^2}}}{{6y + 4}}\]

#### Work Step by Step

\[\begin{gathered}
5{x^3}\,\, = 3{y^2} + 4y \hfill \\
\,Find\,\,dy/dx\,\,\,by\,\,implicit\,\,differentiation \hfill \\
\frac{d}{{dx}}\,\left( {5{x^3}} \right) = \frac{d}{{dx}}\,\left( {3{y^2} + 4y} \right) \hfill \\
by\,\,the\,\,chain\,\,rule \hfill \\
15{x^2} = 6y{y^,} + 4{y^,} \hfill \\
6y{y^,} + 14{y^,} = 15{x^2} \hfill \\
Factor \hfill \\
{y^,}\,\left( {6y + 4} \right) = 15{x^2} \hfill \\
{y^,} = \frac{{15{x^2}}}{{6y + 4}} \hfill \\
Then \hfill \\
{y^,} = \frac{{15{x^2}}}{{6y + 4}} \hfill \\
\end{gathered} \]