The solution of $\frac{dy}{dx}=\frac{y^3+2x^{2}y}{x^3+2x{y}^2}$ is:

Given equation is homogeneous. Let y = vx

$\begin{array}{l}\therefore \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \Rightarrow \frac{y^3+2x^{2}y}{x^3+2x{y}^2}=v+x\frac{dv}{dx}\\ \Rightarrow \frac{(y/x{)}^3+2(y/x)}{1+2(y/x{)}^2}=v+x\frac{dv}{dx}\end{array}$

$\begin{array}{l}\Rightarrow \frac{v^3+2v}{1+2v^2}=v+x\frac{dv}{dx}\\ \Rightarrow x\frac{dv}{dx}=v\left\{\frac{v^2+2}{1+2v^2}-1\right\}=v\left\{\frac{1-v^2}{1+2v^2}\right\}\\ \Rightarrow \frac{1+2v^2}{v\left(1-v^2\right)}dv=\frac{dx}{x}\Rightarrow \frac{1+2v^2}{v(1-v)(1+v)}dv=\frac{dx}{x}\\ \Rightarrow \left(\frac{A}{v}+\frac{B}{1-v}+\frac{D}{1+v}\right)dv=\frac{dx}{x}\end{array}$

where $A(1-v)(1+v)+Bv(1+v)+Dv(1-v)=1+2v^2$
Putting v = 0, A = 1

$\begin{array}{l}v=1,B=\frac{3}{2}\\ v=-1,D=-\frac{3}{2}\\ \therefore \left(\frac{1}{v}+\frac{3}{2}\frac{1}{1-v}-\frac{3}{2}\frac{1}{1+v}\right)dv=\frac{dx}{x}\end{array}$

Integrating both side, we get

$$\begin{gathered} \begin{array}{l}\ln v+\frac{3}{2}\frac{\ln (1-v)}{-1}-\frac{3}{2}\ln (1+v)=\ln x+\ln c\\ \Rightarrow \ln v-\frac{3}{2}\ln (1-v)-\frac{3}{2}\ln (1+v)=\ln cx\\ \Rightarrow v/\left\{\right(1-v)1+v{\}}^{3/2}=cx\\ \Rightarrow {\left(\frac{v}{cx}\right)}^2={\left(1-v^2\right)}^3\\ \Rightarrow {\left(\frac{y}{c{x}^2}\right)}^2={\left(1-\frac{y^2}{x^2}\right)}^3\end{array} \\ \begin{array}{l}\Rightarrow {\left(x^2-y^2\right)}^3=\frac{x^2{y}^2}{c^2}\\ \therefore {\left(x^2-y^2\right)}^3=B{x}^2{y}^2,\left(\because \frac{1}{c^2}=B\right)\end{array} \end{gathered}$$