Jester suggests switching to polar coordinates:
$$x=r\cos(\theta)$$
$$y=r\sin(\theta)$$
and so we find:
$$x^2+y^2=r^2$$
$$\theta=\tan^{-1}\left(\frac{y}{x} \right)$$
$$dx=-r\sin(\theta)\,d\theta+\cos(\theta)\,dr$$
$$dy=r\cos(\theta)\,d\theta+\sin(\theta)\,dr$$
Now, substituting into the ODE, we get:
$$\left(r^3\cos(\theta)- r\sin(\theta) \right)\left(\cos(\theta)\,dr- r\sin(\theta)\,d\theta \right)+ \left(r^3\sin(\theta)+ r\cos(\theta) \right)\left(\sin(\theta)\,dr+ r\cos(\theta)\,d\theta \right)=0$$
Dividing through by $r$ and expanding, we find the first term is:
$$r^2\cos^2(\theta)\,dr-r^3\sin(\theta)\cos(\theta)-\sin(\theta)\cos(\theta)\,dr+r\sin^2(\theta)\,dt$$
and the second term is:
$$r^2\sin^2(\theta)\,dr+r^3\sin(\theta)\cos(\theta)+\sin(\theta)\cos(\theta)\,dr+r\cos^2(\theta)\,dt$$
And so their sum is (and applying the Pythagorean identity):
$$r^2\,dr+r\,d\theta=0$$
Divide through by $r$ to obtain:
$$r\,dr+d\theta=0$$
Integrating, we find:
$$r^2+2\theta=C$$
And back-substituting, we get the general solution:
$$x^2+y^2+2\tan^{-1}\left(\frac{y}{x} \right)=C$$
This is the method I used:
Beginning with:
$$\left(\left(x^2+y^2 \right)x-y \right)\,dx+\left(\left(x^2+y^2 \right)y+x \right)\,dy=0$$
$$\left(x^2+y^2 \right)x-y+\left(\left(x^2+y^2 \right)y+x \right)y'=0$$
Divide through by $$x^2+y^2$$ to get:
$$x+yy'+\frac{xy'-y}{x^2+y^2}=0$$
$$2x+2yy'+2\frac{1}{\left(\frac{y}{x} \right)^2+1}\cdot\frac{xy'-y}{x^2}=0$$
$$2x+2y\frac{dy}{dx}+2\frac{1}{\left(\frac{y}{x} \right)^2+1}\frac{d}{dx}\left(\frac{y}{x} \right)=0$$
Integrating with respect to $x$, we obtain the general solution:
$$x^2+y^2+2\tan^{-1}\left(\frac{y}{x} \right)=C$$