"Let $\omega$ be a 1-form on a smooth manifold $M$ and let $f:M\rightarrow \mathbb{R}$ be a function everywhere non-vanishing such that $d(f\omega)=0$. Prove that $\omega\wedge d\omega=0$".
My idea is: $d(f\omega)=df\wedge \omega+fd\omega=0$ that means $fd\omega=\omega\wedge df$. Now we consider $$f\omega\wedge d\omega=\omega\wedge fd\omega=\omega\wedge\omega\wedge df=0$$ since $\omega\wedge\omega=0$. Now since $f$ is a non-vanishing function, must be $\omega\wedge d\omega=0$. Does it work?