Answer
$$\frac{{dy}}{{dx}} = - \frac{{9\sqrt y }}{{2\sqrt x \left( {12{y^2}\sqrt y - 1} \right)}}$$
Work Step by Step
$$\eqalign{
& 9\sqrt x + 4{y^3} = 2\sqrt y \cr
& 9{x^{1/2}} + 4{y^3} = 2{y^{1/2}} \cr
& {\text{take the derivative on both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {9{x^{1/2}}} \right) + \frac{d}{{dx}}\left( {4{y^3}} \right) = \frac{d}{{dx}}\left( {2{y^{1/2}}} \right) \cr
& {\text{solve the derivatives using the chain rule}} \cr
& 9\left( {\frac{1}{2}} \right){x^{ - 1/2}} + 12{y^2}\frac{{dy}}{{dx}} = 2\left( {\frac{1}{2}{y^{ - 1/2}}} \right)\frac{{dy}}{{dx}} \cr
& {\text{simplifying}} \cr
& \frac{9}{{2\sqrt x }} + 12{y^2}\frac{{dy}}{{dx}} = \frac{1}{{\sqrt y }}\frac{{dy}}{{dx}} \cr
& {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr
& 12{y^2}\frac{{dy}}{{dx}} - \frac{1}{{\sqrt y }}\frac{{dy}}{{dx}} = - \frac{9}{{2\sqrt x }} \cr
& \left( {12{y^2} - \frac{1}{{\sqrt y }}} \right)\frac{{dy}}{{dx}} = - \frac{9}{{2\sqrt x }} \cr
& \left( {\frac{{12{y^2}\sqrt y - 1}}{{\sqrt y }}} \right)\frac{{dy}}{{dx}} = - \frac{9}{{2\sqrt x }} \cr
& \frac{{dy}}{{dx}} = - \frac{{9\sqrt y }}{{2\sqrt x \left( {12{y^2}\sqrt y - 1} \right)}} \cr} $$
