${y}' = \frac{x^{2}+3x^2}{2xy}=\frac{y}{2x}+\frac{3y}{2x} = \frac{1}{2}\left ( \frac{y}{x} \right )^{-1}+\frac{3}{2}\left ( \frac{y}{x} \right )$
$this equation is homogenous $
$u = \frac{y}{x} $
$y = ux$
${y}'={u}'x+u$
${u}'x+u = \frac{1}{2}u^{-1}+\frac{3}{2}u $
seperable
$\left ( \frac{2u}{1+u^{2}} \right )du=\frac{1}{x}dx$
$ln\left |1+u^{2} \right | = ln\left | x \right |$
$1+u^{2}=cx$
$1+\left ( \frac{y}{x} \right )^{2}=cx$
$x^2+y^2= cx^3$
