If $σ$ is an exact differential $1$-form on the plane, then prove that the form $ω=σ+xdy$ is not exact.
In the previous part of the question we have calculated the integral of the differential $ω_1=x \Bbb d y$ over the ellipse $r(t)=\{x=2 \cos t, y=\sin t, \space 0<t<2\pi\}$, giving the answer $2\pi$, so I think it might have something to do with this? But I'm unsure how to start this question.
On
Without using integration, but simply differential calculus: if $\sigma$ is exact and $\omega$ were too, then their difference $\omega - \sigma = x \Bbb d y$ would also be exact. For a trained eye this is clearly not true, but a beginner could use Poincaré's lemma: the interior of the ellipse is contractible (in fact it is convex, and therefore star-shaped) so any exact differential on it should be closed, i.e. $\Bbb d (x \Bbb d y) = 0$, which is equivalent to $\Bbb d x \wedge \Bbb d y = 0$ which is blatantly false.
If $\sigma =d \phi$ is exact, its integral over the ellipse will be zero (by Stokes's theorem: the ellipse bounds a streched-disk $D$ that's a $2$-chain, and $$ \int_{\partial D} d \phi = \int_{D} d^2 \phi = 0. $$
Now you can check: if $\omega$ were exact, its integral over $\partial D$ would also be zero. But because the integral of $x ~dy$ is not zero, you're done.