Answer
$$2{y^3} + 12{y^2} - 3{x^2} + 18x = C$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{x - 3}}{{y\left( {y + 4} \right)}} \cr
& {\text{Separate the variables}} \cr
& y\left( {y + 4} \right)dy = \left( {x - 3} \right)dx \cr
& \left( {{y^2} + 4y} \right)dy = \left( {x - 3} \right)dx \cr
& {\text{Integrate both sides}} \cr
& \int {\left( {{y^2} + 4y} \right)} dy = \int {\left( {x - 3} \right)} dx \cr
& \frac{1}{3}{y^3} + 2{y^2} = \frac{{{x^2}}}{2} - 3x + {C_1} \cr
& 2{y^3} + 12{y^2} = 3{x^2} - 18x + 12{C_1} \cr
& 2{y^3} + 12{y^2} - 3{x^2} + 18x = C \cr} $$
