I guess what I write here is wrong. But, I'm not sure exactly why, so I post it here in the hope someone points out the error of my ways. The two-form $\omega = e^{xy}dx \wedge dy$ may be written as $\omega = e^{r^2 \sin \theta \cos \theta} r \, dr \wedge d\theta$. Then, in polar coordinates, the unit circle is $r=1$ and $\partial / \partial r$ is an outward-pointing vector field on the circle. If I understand the induced orientation of $\omega$ correctly, then all I need to do is to take the inner product (sometimes called "hook") of $\omega$ with $\partial/\partial r$ and pull that back to the circle. Calculate
$$ \omega(\partial/\partial r, \cdot) = e^{r^2 \sin \theta \cos \theta} r dr(\partial/r)d\theta = e^{r^2 \sin \theta \cos \theta} r d\theta$$
but, this is to be pulled-back to the circle. In Cartesians,
$$ d\theta = \frac{-ydx+xdy}{x^2+y^2} $$
thus, using a hybrid notation momentarily,
$$ \omega(\partial/\partial r, \cdot) = e^{r^2 \sin \theta \cos \theta} r \frac{-ydx+xdy}{x^2+y^2} = \frac{-ydx+xdy}{\sqrt{x^2+y^2}}e^{xy}$$
But, I would think the pull-back of this form to the curve $x^2+y^2=1$ simply gives us:
$$ \phi = (-ydx+xdy)e^{xy}$$
Given all this, I'd think $\phi$ is consistent with $dx$ when $y<0$ but, opposite $dx$ when $y>0$. I'm not sure how to get the quoted answers in the stated question. Roughly I follow Lee's Smooth Manifolds, which is also discussed in Inducing orientations on boundary manifolds .