Exterior derivative of differential form

Question

I have problem proving following implication:

Let $\omega=xdy+ydz+zdx$ be a differential form on $\mathbb{R}^3$. If $f : \mathbb{R}^3 \longrightarrow \mathbb{R} $ is a smooth function satisfying $d(f \omega)=0$, then $f=0$.

Using $d(f \omega)=0$ leads to set of equations:

$x\frac{\partial f}{\partial x}-z\frac{\partial f}{\partial y}+f=0$

$y\frac{\partial f}{\partial y}-x\frac{\partial f}{\partial z}+f=0$

$y\frac{\partial f}{\partial x}-z\frac{\partial f}{\partial z}-f=0$

But I don't see that these equations implies that $f=0$. I'd appreciate any help.

Answer 1

Multiply equation (1) by $y$, (2) by $z$ and (3) by $-x$ and add. I reckon you get $(x+y+z)f=0$.