Question:
Let $\alpha$,$\omega$ be $1$-forms of class $C^1$ in $\mathbb R^3$. If $w(x) \neq 0$, for every $x \in \mathbb R^3$ and $\alpha \wedge \omega = 0$. Then $\alpha = f\omega$, where $f : \mathbb R^3 \to \mathbb R$ is a function of class $C^1$.
Attempt: From the lack of innovative ideas, I had to go to the mechanical approach, still not satisfactory though.
Write $\omega = a dx + b dy + c dy $ and $\alpha = r dx + s dy + t dz $ then
$$0 = \omega \wedge \alpha = (as - rb)\, dx \wedge dy + (at - rc)\,dx \wedge dz + (bt - sc)\,dy \wedge dz$$
Thus by linearly independence it follows that $as = rb$ , $at = rc$, $bt = sc$ and from $r = asb^{-1}$ , $t = sb^{-1}c$, $s = sb^{-1}b$ we get $$\begin{align}\alpha &= sb^{-1} a dx + sb^{-1} b dy + sb^{-1}c dz\\&=sb^{-1} (a dx + b dy + c dz)\\&= f \omega\end{align} $$
where $f: \mathbb R^3 \to \mathbb R$ uis given by $f (x) = s \circ b^{-1 }(x)$.
Problems with this approach:
1) Nothing guarantees that such $b^{-1}$ exists since $b : \mathbb R^3 \to \mathbb R$, nor $a^{-1}$,$c^{-1}$ for that matter.
Pointwise, at each point $x\in\mathbb R^3$, $\alpha^{1}\wedge \alpha^{2}\wedge\cdots\wedge \alpha^{k}=0$ if and only if $\{\alpha^i\}$ are linearly dependent.
A proof of the non-trivial direction: if they were independent, they could be chosen as basis forms and evaluating their exterior product at the corresponding dual basis elements we'd have $\alpha^1\wedge\ldots\wedge \alpha^k(v_1,...,v_k)=1.$
In your case $\alpha=f(x)\omega$ and because both 1-forms are required to be of $C^1$ so will be $f(x)$.