Let $M$ be a compact orientable manifold, $\alpha$ and $\beta$ are two volume forms (defined as nowhere vanishing (dim$M$)-forms) on $M$.
Is it true that there exists a smooth function $f:M\rightarrow\mathbb{R}$ such that $\alpha=f\beta$? If it's true, can you prove why? If not, can you please give a counterexample?
This is true, and here is one way of showing it. First consider a chart $(U, x)$ on $M$. There exist smooth functions $f, g : U \to \mathbb{R}$ such that
\begin{align*} \alpha &= g \, dx^1 \wedge \cdots \wedge dx^n \\ \beta &= h \, dx^1 \wedge \cdots \wedge dx^n. \end{align*}
Since, $\alpha, \beta$ are no-where vanishing, so are $g, h$. Thus, $f= \frac{g}{h}$ is well-defined and satisfies that $\alpha = f \beta$ on $U$. Now you can make an argument using a partition of unity and the fact that by compactness you may cover $M$ with finitely many chart domains, the details of which I leave to you, to show that you can make this work globally on $M$.