I'm not sure whether this is acceptable or not, but the simplest thing to do (since the form is closed) is to attempt to integrate it to find supposed $\phi$ such that your form $\omega$ is $\mathrm d \phi$.
One can derive a contradiction most straightforwardly by going around a circle, say $x^2+y^2=1$. We use the path $x(t)=\cos t,y(t)=\sin t$.
$$\int_0^{2\pi} \omega = \int_0^{2\pi} \frac{\cos^2 t +\sin^2 t}{1}\mathrm d t = 2\pi$$
But if $\omega=\mathrm d \phi$ then the expression on the left is $\phi((1,0))-\phi((1,0))=0$, a contradiction.
In response to your complaints about this method: The definition of exact is that $\omega=\mathrm d \phi$ for some $\phi$. The only result we've used is that integrals of derivatives are given by boundary values (a very mini version of Stokes's theorem), i.e.
$$\int_{\mathbf x}^{\mathbf y} \mathrm d \phi = \left[\int_{t_x}^{t_y} \frac{\mathrm d \phi}{\mathrm d t} \mathrm d t = \right]\phi({\mathbf y})-\phi({\mathbf x})$$
which I don't think is really a mysterious result. If there's something you're not happy with do just say.
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, $\beta_F = \star(\alpha_F)$.
Best Answer
Well, I think you want to find nonzero $G$ as $0$ is exact.
Another approach :
De Rham cohomology of $\mathbb{R}^2$ vanishes, thus every closed form is exact. Hence it suffices to show $dG\omega=(G_x dx+G_y dy)\wedge (ydx+dy)-G(dx\wedge dy)=(G_x-yG_y-G)dx\wedge dy=0$. Hence you need to find a function $G$ with $G_x-yG_y-G=0$.