Is this differential form closed / exact? Could you check if I calculated the exterior derivative of this differential form $\omega$ correctly?
$\omega \in \Omega_2 ^{\infty} (\mathbb{R}^3 \setminus \{0\})$
$\omega = (x^2 + y^2 + z^2)^{\frac{-3}{2}}(x \mbox{d}y \wedge \mbox{d}z + y \mbox{d}z \wedge \mbox{d}x + z \mbox{d}x \wedge \mbox{d}y)$
$\mbox{d} \omega = \mbox{d}( (x^2 + y^2 + z^2)^{\frac{-3}{2}}x (\mbox{d}y \wedge \mbox{d}z) + (x^2 + y^2 + z^2)^{\frac{-3}{2}}(y \mbox{d}z \wedge \mbox{d}x ) + (x^2 + y^2 + z^2)^{\frac{-3}{2}}( z \mbox{d}x \wedge \mbox{d}y))$
We differentiate the fist summand only by $x$, second only by $y$ and the third only by $z$, because in the other cases we get forms like $\mbox{d}x \wedge \mbox{d}x$, and due to the fact that exterior derivative is antisymmentric such forms are zero.
So:
$\mbox{d} \omega = \left( - \frac{3}{2} (x^2 + y^2 + z^2)^{\frac{-5}{2}} 2x \cdot x + (x^2 + y^2 + z^2)^{\frac{-3}{2}}\right) \wedge \mbox{d}x \wedge \mbox{d}y \wedge \mbox{d}z + \left( - \frac{3}{2} (x^2 + y^2 + z^2)^{\frac{-5}{2}} 2y \cdot y + (x^2 + y^2 + z^2)^{\frac{-3}{2}}\right) \wedge \mbox{d}y \wedge \mbox{d}z \wedge \mbox{d}x +  \left( - \frac{3}{2} (x^2 + y^2 + z^2)^{\frac{-5}{2}} 2z \cdot z + (x^2 + y^2 + z^2)^{\frac{-3}{2}}\right) \wedge  \mbox{d}z \wedge \mbox{d}x \wedge \mbox{d}y$
Because $\mbox{d}x \wedge \mbox{d}y \wedge \mbox{d}z = \mbox{d}y \wedge dz \wedge dx = \mbox{d}z \wedge \mbox{d}x \wedge \mbox{d}y$ (we need two transpositions), we have:
$\mbox{d} \omega = \left( -3(x^2 + y^2 + z^2) ^{\frac{-5}{2}} (x^2 + y^2 + z^2) + (x^2 + y^2 + z^2) ^{\frac{-3}{2}} \right) \mbox{d}x \wedge \mbox{d}y \wedge \mbox{d}z$
But this isn't equal to zero. Are my calculations correct?
I have one more question - how do determine if this form is exact? I know a differential form $\omega \in \Omega_n (U)$ is exact if there exists $\beta \in \Omega_{n+1} (U)$ s. t. $\omega = \mbox{d} \beta$.
But I don't know how to guess such $\beta$.
Could you help me?
$\mbox{d} \omega = \left( -3(x^2 + y^2 + z^2) ^{\frac{-5}{2}} (x^2 + y^2 + z^2) + (x^2 + y^2 + z^2) ^{\frac{-3}{2}} \right) \mbox{d}x \wedge \mbox{d}y \wedge \mbox{d}z$
Thank you.
EDIT: There is a mistake in my calculations:
There should be 
$\mbox{d} \omega = \left( -3(x^2 + y^2 + z^2) ^{\frac{-5}{2}} (x^2 + y^2 + z^2) + 3(x^2 + y^2 + z^2) ^{\frac{-3}{2}} \right) \mbox{d}x \wedge \mbox{d}y \wedge \mbox{d}z$
and this is zero, so the form is closed.
 A: I did not find any errors in your calculations. So it looks like $\omega$ is not closed.
And how to figure out if you have exact form, given it is closed?
In this case you know that $H^2_{dR}(\mathbb{R}^3 \setminus \{0\}) \simeq \mathbb{R}$ There is just one closed but not exact form(up to scalar multiple). And it is(I'think) $$\alpha = \star d \frac1{\sqrt{x^2+y^2+z^2}}.$$
If you suspect that your form $\omega$ is exact, you know that $\omega + \gamma \alpha$ has to be exact for some $\gamma \in \mathbb{R}$. You can calculate integral $$\int_{S^2} \omega + \gamma \alpha$$ and find $\gamma_0$ for which is the integral zero, than $\omega + \gamma_0 \alpha$ is exact. If $\gamma_0$ happens to be zero than $\omega$ is exact.

edit: Finding $\gamma_0$ is simple thanks to linearity of integral.
$$
\gamma_0 = - \frac{\int_{S^2} \omega}{\int_{S^2} \alpha}
$$
In most cases you just want to show that $\int_{S^2} \omega = 0$. Thanks to that you do not need to know exactly what $\alpha$ is.

edit2: How to compute $\int_{S^2} \omega$
You can think of integrating 2-forms(in 3d) as integrating vector field $\vec F$. 
$$\int \omega = \int \vec F \cdot \vec n dA$$
$\vec n$ is outer normal.
I found this question which discuss correspondence of 1,2-forms and vector fields.
So we can apply this to  $\int_{S^2} \omega$
$$
\int_{S^2} \omega = \int_{S^2} (x^2 + y^2 + z^2)^{\frac{-3}{2}}(x \mbox{d}y \wedge \mbox{d}z + y \mbox{d}z \wedge \mbox{d}x + z \mbox{d}x \wedge \mbox{d}y)=
$$
$$
= \int_{S^2} 1 n\cdot n dA = 4 \pi
$$
because $\vec n = (x,y,z)$ on unit sphere.
