Is the product of a continuously differentiable function with an exact differential exact?

Question

Say you have a function $f$ such that $df$ is exact in an open simply connected set. Is $g\cdot df$ exact, for a differentiable( in that set) function $g$?I think that not necessarily but I can't wrap my head around why. This is inspired in thermodynamics; the differential of entropy is related to the differential of heat as $dS=\dfrac{1}{T} \delta Q$, and $dS$ is perfect. $\delta Q$ isn't supposed to be perfect but if $TdS$ is, then it should be, right? For simplicity, assume $Q$ and $S$ are functions of T only.

Answer 1

Of course not. Take $f(x,y) = xy$ and $df = y\,dx + x\,dy$. Then $\omega = x\,df = xy\,dx + x^2\,dy$ is not exact. You can verify this in at least two ways. Note that $d\omega = (2x-x)\,dx\wedge dy = x\,dx\wedge dy \ne 0$. Or (as in thermodynamics) note that the line (or path) integral is not path-independent. For example, calculate $\int_C\omega$ where $C$ is a path from $(0,0)$ to $(1,1)$. If we go from $(0,0)$ to $(1,0)$ to $(1,1)$, the integral is $1$. If we go from $(0,0)$ to $(0,1)$ to $(1,1)$, then the integral is $1/2$.

COMMENT: The whole point of integrating factors is to go in the other direction. Given $\omega$ with $d\omega\ne 0$ (so $\omega$ cannot be exact), when does there exist a nowhere-zero $f$ so that $f\omega$ is exact? You can find dozens and dozens of posts about this.