Let $M^m$ be a smooth manifold and $f:M\to\mathbb{R}$ a smooth, non singular function. Prove that there exists a vector field $X\in\frak{X}$$(M)$ such that $X(f)\equiv 1$.
Take a chart $(U, \phi)$ at $p\in M$ with $U$ small enough so that we can assume, with no loss in generality, that $\frac{\partial f}{\partial \phi^{n}}\neq 0$ in $U$ (that's possible since $f$ is non singular). That way, take any $g_1, ...,g_{n-1}\in C^{\infty}(M)$ and define
$$g_n:=\left(\frac{\partial f}{\partial \phi^{n}}\right)^{-1}\left(1-\sum_{i=1}^{n-1}g_i\frac{\partial f}{\partial \phi^{i}}\right)$$
which is obviously in $C^{\infty}(M)$. Now, defining $X_{U}:=\sum_{i=1}^n g_i\frac{\partial }{\partial \phi^{i}}$, we have a local smooth vector field with $X_U(f)\equiv 1$ in $U$.
I've tried to use a partition of unity to define some $X:=\sum X_{U}$, but I could not guarantee that $X(f)\equiv 1$ and I don't know how to work this out. Any ideas?