Differential Equations - Exact ODE's 
Hi all, I've attempted the question part b) after rearranging the 1st expression to :
$$v(x,y)\, dx - u(x,y) \, dy = 0.$$
After this I tried using the method of differentiating each expression and trying to calculate if they were exact. It was very lengthy so I am unable to post on here but it seems there is an easier way to go about this. Any help is appreciated.

The equation can be rewritten as $$\frac{1}{(x^2+y^2)^2}\left(2xdx+2ydy\right)y-\frac{2y^2}{(x^2+y^2)^2}dy+\frac{y^2-x^2}{(x^2+y^2)^2}dy+dy=0$$Thus$$y\frac{1}{(x^2+y^2)^2}d(x^2+y^2)+\left(-\frac{1}{x^2+y^2}\right)d(y)+dy=0$$which can be put up as $$y\left\{d\left(-\frac{1}{x^2+y^2}\right)\right\}+\left(-\frac{1}{x^2+y^2}\right)d(y)+dy=0$$Using the product rule of differentiation we have$$d\left\{\frac{-y}{x^2+y^2}\right\}+dy=0$$ So,integrating we get$$-\frac{y}{x^2+y^2}+y=c$$
