#### Answer

\[\frac{{dy}}{{dx}} = \frac{{9{x^2}}}{{16y + 10}}\]

#### Work Step by Step

\[\begin{gathered}
3{x^3} - 8{y^2} = 10y \hfill \\
\,Find\,\,dy/dx\,\,\,by\,\,implicit\,\,differentiation \hfill \\
\frac{d}{{dx}}\,\left( {3{x^3} - 8{y^2}} \right) = \frac{d}{{dx}}\,\left( {10y} \right) \hfill \\
By\,\,the\,\,chain\,\,rule \hfill \\
9{x^2} - 8\,\left( 2 \right)y{y^,} = 10{y^,} \hfill \\
9{x^2} - 16y{y^,} = 10{y^,} \hfill \\
Move\,\,all\,\,{y^,}\,\,terms\,\,to\,\,the\,\,same \hfill \\
- 16y{y^,} - 10{y^,} = - 9{x^2} \hfill \\
Factor \hfill \\
{y^,}\,\left( {16y + 10} \right) = 9{x^2} \hfill \\
{y^,} = \frac{{9{x^2}}}{{16y + 10}} \hfill \\
Then \hfill \\
\frac{{dy}}{{dx}} = \frac{{9{x^2}}}{{16y + 10}} \hfill \\
\end{gathered} \]