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