The general solution of the differential equation \(\left(3 x y+y^2\right) d x\) \(+\left(x^2+x y\right) d y=0\) is

\(
\begin{aligned}
& \left(3 x y+y^2\right) d x+\left(x^2+x y\right) d y=0 \\
& \therefore \frac{d y}{d x}=-\frac{\left(3 x y+y^2\right)}{\left(x^2+x y\right)}
\end{aligned}
\)
Put \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)
\(
\begin{aligned}
& \therefore v+x \frac{d v}{d x}=-\frac{\left(3 v x^2+v^2 x^2\right)}{\left(x^2+v x^2\right)}=-\frac{3 v+v^2}{1+v} \\
& \therefore x \frac{d v}{d x}=-\frac{3 v+v^2}{1+v}-v=-\frac{\left(2 v^2+4 v\right)}{1+v}
\end{aligned}
\)
\(\therefore \int \frac{1+v}{\left(2 v^2+4 v\right)} d v=\int \frac{-1}{x} d x \)
\( \therefore \frac{1}{2} \int \frac{v+1}{v^2+2 v} d v=-\int \frac{d x}{x} \Rightarrow \frac{1}{4} \int \frac{2(v+1)}{v^2+2 v}\) \( d v=-\log x+\log c_1\)
\(\therefore \frac{1}{4} \log \left(\mathrm{v}^2+2 \mathrm{v}\right)+\log \mathrm{x}=\log \mathrm{c}_1 \Rightarrow \frac{1}{4} \log \left(\frac{\mathrm{y}^2}{\mathrm{x}^2}+\frac{2 \mathrm{y}}{\mathrm{x}}\right)\) \(+\log \mathrm{x}=\log \mathrm{c}_1 \)
\( \therefore \frac{1}{4} \log \left(\frac{\mathrm{y}^2+2 \mathrm{xy}}{\mathrm{x}^2}\right)+\log \mathrm{x}=\log \mathrm{c}_1 \Rightarrow \log \left(\frac{\mathrm{y}^2+2 \mathrm{xy}}{\mathrm{x}^2}\right)+\) \(\log \mathrm{x}^4=4 \log \mathrm{c}_1 \)
\( \therefore \log \left[\left(\frac{\mathrm{y}^2+2 \mathrm{xy}}{\mathrm{x}^2}\right)\left(\mathrm{x}^4\right)\right]=\log \mathrm{c}_1^4 \Rightarrow \log \left[\mathrm{x}^2\left(\mathrm{y}^2+2 \mathrm{xy}\right)\right]\) \(=\log \mathrm{c}_1^4 \)
\( \therefore \mathrm{x}^2\left(\mathrm{y}^2+2 \mathrm{xy}\right)=\mathrm{c}\)