differential forms- $\omega $ closed but not exact let be 
$$ \omega= |x|^{-3} \left(x_1 dx_2 \wedge dx_3+x_2dx_3 \wedge dx_1 + x_3dx_1 \wedge dx_2\right) $$ 
and $G:= \mathbb{R}^3 \backslash \{ 0 \} $
I want to prove, that $ \omega$ is closed, but not exact
That $ \omega $ is closed, I can prove it by looking if $ d\omega =0 $
But how can I prove it's not exact?
I know that a continous 2-form is exact in $G$ if there exists a 1-Form $y$ so, that $\omega = dy $
how can I show there doesn't exist such $y$?

Edit:
Showing $\int_S \omega \neq 0 $
Where $S$ is the unitsphere
Set $r=1$ and set
$$x_1= \sin \phi \cos \theta $$
$$x_2= \sin \phi \sin \theta$$
$$x_3= \cos \phi $$
$ \theta=[0, 2\pi], \phi=[0, \pi]$
then $ dx_1 \wedge dx_2 = -\sin \phi \cos \phi d\theta \wedge d\phi $
$ dx_2 \wedge dx_3 = -\sin^2 \phi \cos \theta d\theta \wedge d\phi $
$ dx_3 \wedge dx_1 = -\sin^2 \phi \sin \theta d \theta \wedge d\phi $
Putting in the equation:
$$ \int_0^{2 \pi} \int_0^{ \pi} |x| ( \sin \phi \cos \theta - \sin^2 \phi \cos \theta ~d \theta \wedge d\ \phi \\+ \sin \phi \sin \theta -\sin^2 \phi \sin \theta ~d\theta \wedge d\phi + \cos \phi -\sin \phi \cos \phi~ d \theta \wedge d\phi $$
what do I put for $x$ ?
I don't know how to solve the integral
thank you for any help!

Switch to spherical coordinates and integrate $\omega$ on the unit sphere $r = 1$. Let $\theta,\phi$ denote the azimuth and polar angles, respectively; then $$dx = \cos \theta \sin \phi \, dr - \sin \theta \sin \phi \, d\theta + \cos\theta \cos \phi \, d\phi$$
$$ dy = \sin \theta \sin \phi \, dr + \cos \theta \sin\phi \, d\theta + \sin \theta \cos \phi \, d\phi $$
$$dz = \cos \phi \, dr - \sin\phi \, d\phi$$
and after some algebra we find that $\omega = \sin\phi \, d\phi \wedge d\theta$. Hence $$\int_{S^2} \omega = \int_0^{2\pi} \int_0^\pi \sin \phi \, d\phi \, d\theta = 2\cdot 2\pi = 4\pi.$$
This is the expected answer, namely the surface area of the sphere, as $\omega$ restricted to $S^2$ is precisely the volume form induced by that of $\mathbb R^3$. And since the integral is not zero, $\omega$ cannot be exact, lest Stoke's theorem be violated.
