It can be written in the form of homogeneous equation $\frac{d y}{d x}=-\frac{3 x y+y^2}{x^2+x y}$
So, now put $y=v x$ and $\frac{d y}{d x}-v+x \frac{d v}{d x}$, we get
$\begin{aligned} & \quad v+x \frac{d v}{d x}=-\frac{3 x^2 v+x^2 v^2}{x^2+x^2 v} \Rightarrow x \frac{d v}{d x}=\frac{-2 v(v+2)}{v+1} \\ & \Rightarrow \quad \frac{1}{x} d x=-\frac{v+1}{2 v(v+2)} d v=-\left[\frac{1}{2(v+2)}+\frac{1}{2 v(v+2)}\right] d v \\ & \Rightarrow \quad-\frac{2}{x} d x=\left[\frac{1}{v+2}+\frac{1}{2 v}-\frac{1}{2 v(v+2)}\right]\end{aligned}$
On integrating, we get
$\begin{aligned} & -2 \log _e x=\frac{1}{2} \log (v+2)+\frac{1}{2} \log v+\log c \\ & \Rightarrow v(v+2) x^4=c^2 \Rightarrow \frac{y}{x}\left(\frac{y}{x}+2\right) x^4=c^2 \quad\left(v v=\frac{y}{x}\right)\end{aligned}$
Hence required solution is $\left(y^2+2 x y\right) x^2=c^2$