How to solve the equation-
(x^2+y^2)dx+2xydy=0
you can easily see that this is an exact ODE
$\displaystyle \frac{\partial }
{{\partial y}}\left( {x^2 + y^2 } \right) = \frac{\partial }
{{\partial x}}2xy
$
thus we can find the solution as follows:
$\displaystyle
\psi (x,y) = \int {2xy} dy = xy^2 + f(x)$
we know that:
$\displaystyle
\begin{gathered}
\frac{{\partial \psi (x,y)}}
{{\partial x}} = y^2 + f'(x) = x^2 + y^2 \hfill \\
\Leftrightarrow f(x) = \frac{{x^3 }}
{3} + k \hfill \\
\end{gathered} $
thus the implicit solution is:
$\displaystyle \frac{{x^3 }}
{3} + xy^2 = c$
$\displaystyle \frac{dy}{dx} = -\frac{1}{2} \left(\frac{x}{y} + \frac{y}{x}\right) \Rightarrow x \frac{dv}{dx} = \frac{-3v^2 - 1}{2v}$
(the algebra leading to this should be routine at this level. You do realise that if y = vx then dy/dx is v + x dv/dx, right?).
This DE is seperable: $\displaystyle \frac{dx}{x} = \frac{-2v \, dv}{3v^2 + 1}$.