On a recent differential equations exam I had a task to prove some result involving closed flows and curls, something I have never seen before so I have no idea how to solve it and I hoped I would find some help here.
The question
Let $f,g,\varphi$ be $C^1$ functions from $G$ to $\mathbb{R}$ (where $G$ is a simply connected subset of $\mathbb{R}^2$). If the equation $$C:=\frac{\partial(\varphi f)}{\partial x}+\frac{\partial(\varphi g)}{\partial y}\neq0$$ holds on the whole of $G$, prove that the flow $$x'(t)=f(x,y),\quad y'(t)=g(x,y)$$ doesn't have closed (periodic) orbits (except maybe equilibriums).
My attempts
Generaly, I supposed the opposite: let there be such an orbit, I want to prove that there exists the point where $C=0$. My first attempt was to look for such a point "inside of" the closed orbit (in the area of which the orbit is the boundary), for that I wrongly thought that that whole area is layered in closed orbits and I used linearization in that one equilibrium inside that area. Anyway, that failed. Secondly, I tried looking for that point on the orbit. There, $(f,g)\neq0$ so I tried looking for a strictly positive scalar function $k:G\rightarrow\mathbb{R}$ such that $\frac{\partial(kf)}{\partial x}+\frac{\partial(kg)}{\partial y}=0$ on the orbit. The flow $x'(t)=(kf)(x,y),\;y'(t)=(kg)(x,y)$ has the same orbits as the original flow. Then, whereever $\varphi/k$ reaches an extremum on the orbit we have that $C=0$ (it could be directly checked). But I couldn't find such $k$.