Answer
$$\frac{{dy}}{{dx}} = \frac{y}{{4y - x}}$$
Work Step by Step
$$\eqalign{
& \frac{{xy}}{2} - {y^2} = 3;{\text{ }}\frac{{dy}}{{dx}} \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \underbrace {\frac{d}{{dx}}\left[ {\frac{{xy}}{2}} \right]}_{{\text{Use product rule}}} - \underbrace {\frac{d}{{dx}}\left[ {{y^2}} \right]}_{{\text{Use chain rule}}} = \frac{d}{{dx}}\left[ 3 \right] \cr
& \frac{x}{2}\frac{d}{{dx}}\left[ y \right] + \frac{y}{2}\frac{d}{{dx}}\left[ x \right] - {y^{2 - 1}}\frac{{dy}}{{dx}} = 0 \cr
& \frac{1}{2}x\frac{{dy}}{{dx}} + \frac{1}{2}y - 2y\frac{{dy}}{{dx}} = 0 \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& \frac{1}{2}x\frac{{dy}}{{dx}} - 2y\frac{{dy}}{{dx}} = - \frac{1}{2}y \cr
& \left( {\frac{1}{2}x - 2y} \right)\frac{{dy}}{{dx}} = - \frac{1}{2}y \cr
& \left( {x - 4y} \right)\frac{{dy}}{{dx}} = - y \cr
& \frac{{dy}}{{dx}} = - \frac{y}{{x - 4y}} \cr
& \frac{{dy}}{{dx}} = \frac{y}{{4y - x}} \cr} $$
