Differential form is exact, but vector field is not conservative - what did I miss? I have the following differential form $\omega\in\Omega_2(\mathbb{R^3})$:
$$\omega = 2xzdy\wedge dz + dz\wedge dx - (z^2+e^x)dx\wedge dy$$
The question is whether it is exact, i.e whether there exists $\eta\in\Omega_1(\mathbb{R^3})$ such that $\omega = d\eta$.
Based on this definition, $\omega$ is indeed exact. I have found $\eta = zdx - (z^2x-e^x)dy$ that works. 
However, I also know that $\omega$ exact $\Leftrightarrow$ $F$ conservative, where $F$ is a vector field associated with $\omega$: $F=(2xz,1,-(z^2+e^x))$. And $F$ is conservative and the domain of $F$ is simply connected $\Leftrightarrow$ curl of $F$ is zero.
From this follows: $\omega$ exact $\Leftrightarrow$ $F$ conservative $\Leftrightarrow$ curl of $F$ is zero. Thus checking for curl of $F$ is a simple test for checking for a differential form being exact. But in this case, the curl of $F$ is $(0,2x+e^x,0) \neq (0,0,0)$. So the vector field is not conservative. But we know that $\omega$ is exact, so $F$ must be conservative. And here I have arrived at contradiction. Do I miss something here?   
 A: You usually associate a vector field $F$ to a 1-form $\varphi$. Then, assuming that your space is simply-connected, 
$\mathrm{curl}\,F = 0 \Leftrightarrow d\varphi = 0\Leftrightarrow \varphi = d\alpha$ for some scalar function $\alpha$. 
Your $\omega$ is a 2-form, so $d\omega$ is a 3-form. You can also associate a vector field to this, but then you require $\mathrm{div}\,F = 0$. (However you need additional topological information – there can be a sphere that is not shrinkable).
A: Note that there is no contradiction anywhere; the form $\omega$ you defined is exact (but $\eta$ has a typo it should be $\eta = z \, dx - (z^2 x + e^x) \, dy$), but your intermediate reasoning is false.
Let's explore why: in $\Bbb{R}^3$ specifically, there are two ways to relate vector fields and differential forms.  Let $F: \Bbb{R}^3 \to \Bbb{R}^3$ be a given vector field. The first is to define a $1$-form $\alpha_F$, defined as:
\begin{align}
\alpha_F := F_1 \,dx + F_2 \, dy + F_3 \, dz
\end{align}
A second way is to define a $2$-form $\beta_F$ defined as:
\begin{align}
\beta_F := F_1\, dy \wedge dz + F_2 \, dz \wedge dx + F_3 \, dx \wedge dy
\end{align}
Now, we can calculate the exterior derivative of both these forms, and after a few lines, you'll find that:
\begin{align}
\begin{cases}
d(\alpha_F) &= \beta_{\text{curl}(F)} \\\\
d(\beta_F) &= \text{div} (F) \, dx \wedge dy \wedge dz
\end{cases}
\end{align}
So, the correct equivalences are that since $\Bbb{R}^3$ is star-shaped,
\begin{align}
\text{$\alpha_F$ is exact} & \iff \text{ $\alpha_F$ is closed} \\
& \iff d(\alpha_F) = \beta_{\text{curl}(F)} = 0 \\
& \iff \text{curl}(F) = 0
\end{align}
and
\begin{align}
\text{$\beta_F$ is exact} & \iff \text{ $\beta_F$ is closed} \\
& \iff d(\beta_F)  = \text{div} (F) \, dx \wedge dy \wedge dz = 0 \\
& \iff \text{div}(F) = 0
\end{align}
You said:

Thus checking for curl of $F$ is a simple test for checking for a differential form being exact.

Well, that's only true if you're working with the first type of form, like $\alpha$. But in your question, you're working with the second type, $\beta$. So, in your case, the "simple" way to check whether your form $\omega$ is exact is to see if the divergence of $F$ vanishes; and indeed the divergence of $F(x,y,z) = (2xz, 1, - (z^2 + e^x))$ is $0$. Thus, the form $\omega$ you have is exact.

By the way, the general framework for the above constructions is the following. If you want to be slightly fancy, you can say that on $\Bbb{R}^3$, we have a "standard" Riemannian metric tensor field $g = dx \otimes dx + dy \otimes dy + dz \otimes dz$, so that given a vector field $F = F_1 \dfrac{\partial }{\partial x} + F_2 \dfrac{\partial }{\partial y} + F_3 \dfrac{\partial }{\partial z}$, we can use the musical isomorphism to get a $1$-form, $\alpha_F := g^{\flat}(F)$. Also, in $\Bbb{R}^3$, we can provide an orientation so that we can define the Hodge-star operator, which in general sends $k$-forms to $n$-k forms (so in $n = 3$ dimensions, it sends $1$-forms to $2$-forms). In this case, $\beta_F = \star(\alpha_F)$.
