Let $d(fw)=0$ for $f\neq 0$ show that $w \wedge dw =0$ where w is a 1-form.

Question

Let $d(fw)=0$ for $f\neq 0$ show that $w \wedge dw =0$ where w is a 1-form and f is a 0-form

My attempt:

Let $c=1/f$ then $cw \wedge d(fw)=0$ implies $cw \wedge (df \wedge w+f \wedge dw)=0$

implies $cw \wedge df \wedge w + cw \wedge f \wedge dw=0$

implies $cw \wedge df \wedge w + cw \wedge f dw=0$

implies $cw \wedge df \wedge w - cfdw \wedge w=0$

implies $cw \wedge df \wedge w + w \wedge dw=0$

and then I'm stuck, I'm trying to cancel out the left term and get my result, it's not working so well though.

Any help would really be appreciate, thank you so much

Answer 1

$d(fw)=0$ implies that $df\wedge w +f.dw =0$ implies $df\wedge w\wedge w +f.dw\wedge w =0$ and clearly $w\wedge w =0$ So, $f.dw\wedge w =0$ & we have $f\neq 0$. So $dw\wedge w =0$.