How about doing this
$\displaystyle \frac{x dx}{(x^2+y^2)^{3/2}} = - \frac{y dy}{(x^2+y^2)^{3/2}} $
multiply both sides by $\displaystyle (x^2+y^2)^{3/2} $
to get
$\displaystyle x dx = - y dy $
or
$\displaystyle - \frac{x^2}{2} + c = \frac{y^2}{2} $