Solve $x^2(xdx+ydy)+2y(xdy-ydx)=0$ Solve $x^2(xdx+ydy)+2y(xdy-ydx)=0$
My Attempts 
$$x^2(xdx+ydy)+2y(xdy-ydx)=0$$
$$\dfrac {xdx+ydy}{xdy-ydx}=-\dfrac {2y}{x}$$
Put $x=r\cos (\theta)$ and $y=r\sin (\theta)$
So, $r^2=x^2+y^2$ and $\tan (\theta)=\dfrac {y}{x}$
Now,
$$x^2+y^2=r^2$$
Differentiating both sides,
$$2xdx+2ydy=2rdr$$
$$xdx+ydy=rdr$$

$$\dfrac {xdx+ydy}{xdy-ydx}=-\dfrac {2y}{x}$$
Duvide both sides by $({x^2+y^2})$:
$$\dfrac {xdx+ydy}{x^2+y^2}=-\dfrac {2y}{x}\left ( \dfrac {xdy-ydx}{x^2+y^2} \right )$$
$$\dfrac {xdx+ydy}{x^2+y^2}=-\dfrac {2y}{x}\left ( d(\arctan (\frac {y}{x})) \right )$$
Susbtitute $\dfrac {y}{x}=z$
$$\frac 12\dfrac {d(x^2+y^2)}{x^2+y^2}=-2zd(\arctan z)$$
$$\frac 12\dfrac {d(x^2+y^2)}{x^2+y^2}=-\dfrac {2zdz}{z^2+1}$$
Integrate.

Edit I didn't pay attention that the right side of the DE was wrong so:
$$\frac 12\dfrac {d(x^2+y^2)}{x^2+y^2}=-2\frac {y}{\color{red}{x^2}}d(\arctan \frac {y}{x})$$
$$\frac 12\dfrac {dr^2}{r^2}=-2\frac {r\sin \theta}{r^2 \cos^2 \theta}d\theta$$
More simply:
$${dr}=-2\frac {\sin \theta}{\cos^2 \theta}d\theta$$
Integrate.
