Consider the smooth differential 2-form $\omega = x \ dy \wedge dz$ defined on $\mathbb{R}^{3}$ (using the coordinates $(x,y,z)$ of course).
I'm trying to show that $\omega$ is NOT exact on any open subset of $\mathbb{R}^{3}$.
Suppose that $\omega$ is exact on some open subset $A \subseteq \mathbb{R}^{3}$. By definition, then there must exist a smooth differential $1$-form $\eta$ such that $d \eta = \omega$.
Since $\eta$ is a smooth differential 1-form, there exist functions $f,g,h \in C^{\infty}(A)$ such that:
$\eta = f \ dx + g \ dy + h \ dz$
Taking the differential of $\eta$, and computing a little, I find that:
$d \eta = \left[ \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \right] dx \wedge dy + \left[ \frac{\partial h}{\partial x} - \frac{\partial f}{\partial z} \right] dx \wedge dz + \left[ \frac{\partial h}{\partial y} - \frac{\partial g}{\partial z} \right] dy \wedge dz$
But since $d\eta = \omega$, this means that:
$\begin{cases} \ \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} = 0 \\ \ \frac{\partial h}{\partial x} - \frac{\partial f}{\partial z} = 0 \\ \ \frac{\partial h}{\partial y} - \frac{\partial g}{\partial z} = x \end{cases}$
....I get very stuck here. I am guessing there is some way to show that the above system of PDEs has no solution, and hence $\omega$ is NOT exact. Where do I go from here?
As noted in the comments, this is very simple to answer; but just for interest here's how to make your approach work:
Differentiating the second equation by $y$ gives $h_{xy} = f_{zy}$, and the first by $z$ gives $g_{xz} = f_{zy}$. Thus we have $h_{xy} = g_{xz}$. But differentiate the third equation by $x$ and we get $h_{xy} = g_{xz} + 1$, a contradiction.