Answer
$C=x^2y+y^2$
Work Step by Step
We are given:
\[\frac{dy}{dx}=-\frac{2xy}{x^2+2y}\]
$$(x^2+2y)dy=-2xydx=0$$
We have:
$$M(x,y)=2xy \wedge N(x,y)=x^2+2y$$
Check:
$$M'(x,y)=2x = N'(x,y)$$
The differential equation is exist.
Let's set:
$$\phi (x,y)=0$$
There exists a potential function $phi$ satisfies:
$$\frac{\partial \phi}{\partial x}=M \wedge \frac{\partial \phi}{\partial y}=N$$
Integrating $\frac{\partial \phi}{\partial x}=2xy$ we get:
$$\phi = x^2y+f(y)$$
Since $\frac{\partial \phi}{\partial y}=x^2+f'(y)=x^2+2y$
We get:
$$f'(y)=2y \rightarrow f(y)=y^2+C$$
$C$ is constant of integration
Yield:
$$\phi=x^2y+y^2+C$$
Hence general solution is $x^2y+y^2=C$
