Official solution
(x2 - y2) dx + 2xy dy = 0
\[{{dy} \over {dx}} = {{{y^2} - {x^2}} \over {2xy}}\]
Put \[y = vx \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}\]
Solving we get,
\[\int {{{2v} \over {{v^2} + 1}}dv = \int { - {{dx} \over x}} } \]
ln(v2 + 1) = - ln x + C
(y2 + x2) = Cx
1 + 1 = C \[ \Rightarrow \] C = 2
y2 + x2 = 2x