If $\omega$ is an exact differential form on $\Bbb R^2$, and $c_1$ and $c_2$ are two arcs with the same starting/ending points, then
$$
\int_{c_1} \omega = \int_{c_2} \omega.
$$
With that in mind, consider the two paths from $(+1, 0)$ to $(-1, 0)$ that go, respectively, around the top semicircle and the bottom semicircle of the unit circle.
Let me amplify a little bit, because this problem, in more or less this form, reappears constantly, and it's a pity it's not stated better.
There's a function $\theta(x, y) = arctan(y/x)$, defined for points $(x, y)$ in the plane with $x \ne 0$. Let's denote by
$U$ the set $\{ (x, y) \mid x \ne 0 \}$, so $U$ is the domain of $\theta$.
The exterior derivative of $\theta$ is also defined on $U$ (because $\theta$ is differentiable). A little computation shows that this exterior derivative has the formula (for points $(x, y)$ that are actually in $U$)
\begin{align}
d\theta (x, y)
&= \frac{\partial \theta} {\partial x}(x, y) dx +
\frac{\partial \theta} {\partial y}(x, y) dy \\
&= \frac{1} {1 + (y/x)^2}(\frac{-y}{x^2}) dx +
\frac{1}{1 + (y/x)^2} \frac{1}{x} dy \\
&= \frac{-y} {x^2 + y^2} dx +
\frac{x}{x^2 + y^2} dy.
\end{align}
This 1-form is defined on $U$, and is an exact $1$-form on $U$; it's the exterior derivative of the function $\theta$, after all!
But algebraically, if you look at it, it makes sense for any pair of numbers $(x, y)$ except $(x, y) = (0, 0)$; in other words, it admits a continuous extension to the set
$$
V = \Bbb R^2 - \{ (0,0) \}.
$$
Indeed, because $U$ is a dense subset of $V$, there's at most one continuous extension to $V$, and hence it must be the one given by the algebraic formula above.
Now here's the stupid part: even though the function $\theta$ is defined on $U$ rather than $V$, and hence $d\theta$ is only defined on $U$, this unique continuous extension to $V$ is also called "$d\theta$". I can't fix that, but I can ignore it for now, and I'm going to. I'm going to define a $1$-form on $V$ called $\eta$, by the formula:
\begin{align}
\eta (x, y)
&= \frac{-y} {x^2 + y^2} dx +
\frac{x}{x^2 + y^2} dy.
\end{align}
You'll notice that (1) $\eta$ is a nice smooth 1-form. (2) On the subset $U$, we have $\eta(x, y) = d\theta(x, y)$.
Because $S^1$ is a subset of $V$, we can restrict $\eta$ to be a 1-form on $S^1$ as well, if we like.
The thing that the problem is asking you to show (hard though it may be to believe, given that they left out all this stuff) is that there's no 0-form (i.e., differentiable function) $f$ on either $V$ or $S^1$, with the property that $df = \eta$.
There are two approaches you can take.
The first is to use the observation I made above, and note that the integral of $\eta$ over the upper and lower semicircle paths from $(+1, 0)$ to $(-1, 0)$ are different --- one is $+\pi$; the other is $-\pi$.
The second is to note that if such a function $f$ exists on $V$, then on the right open half-plane, which I'll call $V^+$, it must be the same as $\theta$, up to an additive constant, $c_1$ because they have the same partial derivatives; the same is true on the left half-plane, $V^-$, up to a possibly different additive constant $c_2$. So we would have, on $V^+$, that
$$
f(x, y) = \theta(x, y) + c_1.
$$
Fixing $y = 1$ and letting $x$ approach $0$ from above, we'd be taking arctan of large positive numbers, and we'd find that
$$
f(0, 1) = \frac{\pi}{2} + c_1.
$$
On the other hand, fixing $y = 1$ and letting $x$ approach $0$ from below, we're taking $\arctan$ of large negative numbers, and conclude that
$$
f(0, 1) = -\frac{\pi}{2} + c_2.
$$
so that $c_2 - c_1 = \pi$.
Doing the same analysis at the point $(0, -1)$, we would find that $c_1 - c_2 = \pi$. And that's a contradiction.
Best Answer
Just use Cartan's magic formula: $$\begin{align}\mathcal{L}_{\sin x\,\partial_y} (\sin x\,{\rm d}y) &= \iota_{\sin x\,\partial_y}({\rm d}(\sin x\,{\rm d}y)) + {\rm d}(\iota_{\sin x\,\partial_y}(\sin x\,{\rm d}y)) \\ &= \iota _{\sin x\,\partial_y}(\cos x\,{\rm d}x\wedge {\rm d}y) + {\rm d}(\sin^2x) \\ &=\cos x\begin{vmatrix}0 & {\rm d}x \\ \sin x & {\rm d}y\end{vmatrix} +2\sin x\cos x\,{\rm d}x \\ &= - \sin x \cos x\,{\rm d}x +2\sin x\cos x\,{\rm d}x \\ &= \sin x\cos x\,{\rm d}x .\end{align}$$ I'm not sure what happened in your calculation. It looked like you wrote $\mathcal{L}_{\partial_y}({\rm d}y)=1$, but this should be a $1$-form and not a function. Since $\mathcal{L}_X$ is a derivation, you would have that $$\mathcal{L}_X({\rm d}y)(Y) = X({\rm d}y (Y)) - {\rm d}y([X,Y]).$$I reccomend the Analysis and Algebra on Differentiable Manifolds workbook. There's a lot of computations we normally don't see carried out in the classical references.