checking if a 2-form is exact Consider the 2-form 
$$\sigma=\frac{x_1 dx_2 \wedge dx_3 + x_2dx_3\wedge dx_1+ x_3 dx_1 \wedge dx_2}{(x_1^2+x_2^2+x_3^2)^{3/2}}.$$
I need to show if it is exact or not. Suppose it is exact, then there exists a 1-form
$$\alpha=f_1dx_1+f_2dx_2+f_3dx_3,$$
such that 
$$d\alpha=\displaystyle\sum_i df_i\wedge dx_i=\sigma.$$
Now I use $$df_i=\frac{\partial f_i}{\partial x_1}dx_1+\frac{\partial f_i}{\partial x_2}dx_2+\frac{\partial f_i}{\partial x_3}dx_3,$$
then $d\alpha$ becomes
$$\left(\frac{\partial f_2}{\partial x_1}-\frac{\partial f_1}{\partial x_2}\right)dx_1\wedge dx_2+\left(\frac{\partial f_3}{\partial x_2}-\frac{\partial f_2}{\partial x_3}\right)dx_2\wedge dx_3+\left(\frac{\partial f_1}{\partial x_3}-\frac{\partial f_3}{\partial x_1}\right)dx_3\wedge dx_1.$$
I do not know how to proceed, is this even the right method?
 A: Generally it's easier to prove that a form is NOT exact.
For instance, an exact form is necessarily closed. So, if your form was exact it would be closed. And to verify this later property is just a matter of differentiation -which is easier than looking for an antiderivative, as you seem to be trying. Of course, if $d\sigma = 0$, this says nothing about being exact or not.
Another idea: if your form was exact, $\sigma = d\alpha$, for some $\alpha$, then, by Stokes' theorem, its integral over a closed surface $S$ (that is, with no boundary, or empty boundary, $\partial S = \emptyset$) would be zero:
$$
\int_S d\alpha = \int_{\partial S} \alpha = 0 \ .
$$
So you could try to find a closed surface $S$ such that $\int_S \sigma \neq 0$.
Where to look for such a surface? Well, there is a "meta criterion" for this: since this is, presumably, an exercise in a beginner's differential geometry book, it can NOT be too far away from your knowledge.  :-)  So, I would try with the first closed surface that came to my mind without hesitation.
Or also, looking at that denominator, you could think of a surface where it becomes really, really, and I mean really, simple. (For instance, an sphere of radius 1.)
A: It's not exact. To see this, as suggested by Andrea, we use the spherical coordinates: 
$$(x_1,x_2,x_3)=(r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi),0<\theta<2\pi,0<\phi<\pi.$$
Restricted to $\mathbb{S}^2$, we have $r=1$, i.e. 
$$(x_1,x_2,x_3)=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi),$$
which implies that
$$dx_1|_{\mathbb{S}^2}=-\sin\phi\sin\theta\,d\theta+\cos\phi\cos\theta\,d\phi,$$
$$dx_2|_{\mathbb{S}^2}=\sin\phi\cos\theta\,d\theta+\cos\phi\sin\theta\,d\phi,$$
$$dx_3|_{\mathbb{S}^2}=-\sin\phi\,d\phi.$$
This gives
$$dx_1\wedge dx_2\big|_{\mathbb{S}^2}=-\sin\phi\cos\phi\,d\theta\wedge d\phi,$$
$$dx_2\wedge dx_3\big|_{\mathbb{S}^2}=-\sin^2\phi\cos\theta\,d\theta\wedge d\phi,$$
$$dx_3\wedge dx_1\big|_{\mathbb{S}^2}=-\sin^2\phi\sin\theta\,d\theta\wedge d\phi.$$
Combining all these, we can express $\sigma$ in terms of spherical coordinates:
$$\sigma\big|_{\mathbb{S}^2}=\Big(\frac{x_1 dx_2 \wedge dx_3 + x_2dx_3\wedge dx_1+ x_3 dx_1 \wedge dx_2}{(x_1^2+x_2^2+x_3^2)^{3/2}}\Big)\Big|_{\mathbb{S}^2}$$
$$=\big(x_1 dx_2 \wedge dx_3 + x_2dx_3\wedge dx_1+ x_3 dx_1 \wedge dx_2\big)\big|_{\mathbb{S}^2}$$
$$=-\sin^3\phi\cos^2\theta\,d\theta\wedge d\phi-\sin^3\phi\sin^2\theta\,d\theta\wedge d\phi-\sin\phi\cos^2\phi\,d\theta\wedge d\phi$$
$$=-\sin\phi\,d\theta\wedge d\phi.$$
Suppose that $\sigma$ is  exact, then $\sigma=df$ for some one-form $f$. By Stoke's theorem, we have
$$\int_{\mathbb{S}^2}\sigma=\int_{\mathbb{S}^2}df=\int_{\partial \mathbb{S}^2}f=0$$
where the last equality follows from the fact that $\partial \mathbb{S}^2=\varnothing$, i.e. $\mathbb{S}^2$ is an manifold with no boundary. However, from the above calculation, 
$$\int_{\mathbb{S}^2}\sigma=-\int_{0}^\pi\int_0^{2\pi}\sin\phi\,d\theta d\phi=-4\pi, $$
which is a contradiction. 
