Answer
$$dy = \frac{{6\sqrt x - {y^2}}}{{2xy - 1}}dx$$
Work Step by Step
$$\eqalign{
& x{y^2} - 4{x^{3/2}} - y = 0 \cr
& {\text{Calculate }}\frac{{dy}}{{dx}}{\text{ by implicit differentiation}} \cr
& \frac{d}{{dx}}\left[ {x{y^2}} \right] - \frac{d}{{dx}}\left[ {4{x^{3/2}}} \right] - \frac{d}{{dx}}\left[ y \right] = 0 \cr
& x\frac{d}{{dx}}\left[ {{y^2}} \right] + {y^2}\left( 1 \right) - 6{x^{1/2}} - \frac{{dy}}{{dx}} = 0 \cr
& 2xy\frac{{dy}}{{dx}} + {y^2} - 6\sqrt x - \frac{{dy}}{{dx}} = 0 \cr
& 2xy\frac{{dy}}{{dx}} - \frac{{dy}}{{dx}} = 6\sqrt x - {y^2} \cr
& \left( {2xy - 1} \right)\frac{{dy}}{{dx}} = 6\sqrt x - {y^2} \cr
& \frac{{dy}}{{dx}} = \frac{{6\sqrt x - {y^2}}}{{2xy - 1}} \cr
& {\text{Write in differential form}} \cr
& dy = \frac{{6\sqrt x - {y^2}}}{{2xy - 1}}dx \cr} $$
