Answer
$$\frac{{dy}}{{dx}} = \frac{{2\sqrt y }}{{\sqrt x \left( {9y + 4} \right)}}$$
Work Step by Step
$$\eqalign{
& 4\sqrt x - 8\sqrt y = 6{y^{3/2}} \cr
& {\text{take the derivative on both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {4\sqrt x - 8\sqrt y } \right) = \frac{d}{{dx}}\left( {6{y^{3/2}}} \right) \cr
& \frac{d}{{dx}}\left( {4\sqrt x } \right) - \frac{d}{{dx}}\left( {8\sqrt y } \right) = \frac{d}{{dx}}\left( {6{y^{3/2}}} \right) \cr
& {\text{solve the derivatives }} \cr
& 4\left( {\frac{1}{{2\sqrt x }}} \right) - 8\left( {\frac{1}{{2\sqrt y }}} \right)\frac{{dy}}{{dx}} = 6\left( {\frac{3}{2}} \right){y^{1/2}}\frac{{dy}}{{dx}} \cr
& \frac{2}{{\sqrt x }} - \frac{4}{{\sqrt y }}\frac{{dy}}{{dx}} = 9{y^{1/2}}\frac{{dy}}{{dx}} \cr
& {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr
& \frac{2}{{\sqrt x }} = 9\sqrt y \frac{{dy}}{{dx}} + \frac{4}{{\sqrt y }}\frac{{dy}}{{dx}} \cr
& \frac{2}{{\sqrt x }} = \left( {9\sqrt y + \frac{4}{{\sqrt y }}} \right)\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{2}{{\sqrt x }}}}{{9\sqrt y + \frac{4}{{\sqrt y }}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{{2\sqrt y }}{{\sqrt x }}}}{{9y + 4}} \cr
& \frac{{dy}}{{dx}} = \frac{{2\sqrt y }}{{\sqrt x \left( {9y + 4} \right)}} \cr} $$
