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.)
Best Answer
Hint Define $\gamma : \mathbb{R} \to \mathbb{R}^4$ to be the solution of $\gamma' = X(\gamma)$ and $\gamma(0) = p \in \mathbb{R}^4$. The flow you want to compute is $\varphi_t^X(p) = \gamma(t)$.
Write $\gamma = (x_0(t),x_1(t),x_2(t),x_4(t))$ in coordinates and find the differential equations verified by the coordinates. For example, for the first and second coordinate: \begin{align} {x_0}'(t) &= -\partial_1u(x_0(t),x_1(t),x_2(t),x_3(t))\\ {x_1}'(t) &= \partial_0 u(x_0(t),x_1(t),x_2(t),x_3(t)) \end{align}
Remark it appears constant are not integral curves of the vector field if $\partial u$ is not zero.
Edit To answer the (new) question there is no need to compute the flow, but just to use its definition. By the very definition of the Lie derivative of a differential form in the direction of a vector field, one has \begin{align} \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\varphi^*_t\omega = \mathcal{L}_X\omega \end{align} By the Cartan's (magic) formula, \begin{align} \mathcal{L}_X\omega = \mathrm{d}\left(\omega(X,\cdot)\right) + \mathrm{d}\omega\left(X,\cdot\right) \end{align} As $\mathrm{d}\omega = 0$ (easy computation), it is \begin{align} \mathcal{L}_X\omega = \mathrm{d}\left(\omega(X,\cdot)\right) \end{align} But \begin{align} \omega(X,\cdot) &= -\partial_0u\mathrm{d}x^0 -\partial_1u\mathrm{d}x^1-\partial_2u\mathrm{d}x^2-\partial_3u\mathrm{d}x^3 \\ &= -\mathrm{d}u \end{align} and thus \begin{align} \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\varphi^*_t\omega = \mathrm{d}\left(-\mathrm{d}u\right) = -\mathrm{d}^2u = 0 \end{align} and consequently, $\varphi_t^*\omega$ is constant. As $\varphi_0 = \mathrm{id}$, it is equal to $\omega$.