I want to prove the following:
Show that $H^2(\mathbb R^3 \setminus \{0\})$ (the second de Rham cohomology space) does not vanish. Hint: Use the form $\omega = i_V \lambda_3$ with the standard volume form $\lambda_3 = dx_1 \wedge dx_2 \wedge dx_3$ and the vector field $V: \mathbb R^3 \setminus \{0\} \to \mathbb R^3$, $V(x) := x/ \lVert x \rVert_2^3$.
I already proved that $d\omega = 0$ using $i_V\lambda_3 = x_1/ \lVert x \rVert_2^3 dx_2 \wedge dx_3 - x_2/ \lVert x \rVert_2^3 dx_1 \wedge dx_3 + x_3/ \lVert x \rVert_2^3 dx_1 \wedge dx_2$
and then calculating the exterior derivative (is there a faster way?).
Now I want to calculate $$\int_{S^2}\omega$$ but I don't know how to do that. The interior product $i_V$ really confuses me here. Any help appreciated.
The faster way to see $\omega$ is closed is to pull back by a spherical coordinate parametrization of $\Bbb R^3-\{0\}$. With $g(\rho,\phi,\theta) = \rho(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)$, you can easily check that $g^*\omega = \sin\phi\,d\phi\wedge d\theta$. Of course, $g^*(d\omega) = dg^*\omega = 0$.
Indeed, this then gives you an easy way to compute $\displaystyle\int_{S^2}\omega$, since the unit sphere is given precisely by $\rho=1$. Just parametrize (almost all) the sphere by $(0,\pi)\times (0,2\pi)$ in $\phi\theta$-space.
Alternatively, note from your formula that on the unit sphere we have $$\omega = x_1\,dx_2\wedge dx_3 + x_2\,dx_3\wedge dx_1+x_3\,dx_1\wedge dx_2,$$ and this is precisely the (standard) area $2$-form on the sphere (because — again on the unit sphere — $V$ is the outward-pointing unit normal vector to the sphere).