Given $\omega = fdx+gdy+hdz$, such that $\omega\wedge dz=0$, what can we conclude about $f$, $g$ and $h$?
Let's write what we have:
$$ 0 = \omega\wedge dz = f dx\wedge dz + g dy \wedge dz. $$
We could take $d$ here and get $g_x = f_y$. Then we could use this result and $d^2 \omega = 0$ and obtain $f_{zy}=g_{zx}$.
So I derived something, but not really understand the geometric sense or what this all was about. But it should be smth.
The 2-forms $dx\wedge dy, dy\wedge dz, dz\wedge dx$ are a basis, in particular
$dx\wedge dz$ and $dy\wedge dz$ are independent.
So, when you have $0 = f dx\wedge dz + g dy\wedge dz$, then $f$ and $g$ must both be zero.
Here is a geometric interpretation: evaluating $\omega\wedge dz (\vec u,\vec v)$ at a given point, $p$, is proportional to the following area:
project the parallelogram spanned by $\vec u, \vec v$ onto the plane spanned by $$\{\vec\omega_p := (f(p),g(p),h(p)), \partial_z := (0,0,1)\}$$ and find its oriented area. (if you like formulas, you can show $\omega\wedge dz (\vec u, \vec v) = (\vec\omega_p\times \partial_z) \cdot (\vec u\times \vec v)$)
If this is always zero, then $\vec\omega_p$ and $\partial_z$ are dependent (do not span a plane), i.e. $\vec\omega_p$ is a multiple of $\partial_z$ or $f(p) = g(p) = 0$ (at an arbitrary point $p$).