Using the invariant formula for the exterior derivative, one gets that for a $3$-form $\omega$ its derivative is evaluated on vector fields according to
$$3 \ {\rm d} \omega (X, Y, Z) = X \ \omega (Y, Z) + Y \ \omega (Z, X) + Z \ \omega (X, Y) - \omega ( [X, Y], Z) - \omega ( [Y, Z], X) - \omega ( [Z, X], Y) .$$
This is supposed to be a tensor expression, but I'm having troubles verifying this. If I replace $X$ by $fX$, with $f$ a smooth function, instead of $f$ simply getting out and multiplying everything, I get the following terms containing derivatives of $f$: the second term produces $+Yf$, the third $+Zf$, the fourth $+Yf$ and the sixth $-Zf$.
Why don't the terms with $Yf$ cancel out like the ones with $Zf$? The same happens if I multiply $Y$ or $Z$ by $f$.
They do cancel, if you keep track of the signs correctly. For example, the second term produces $(Yf) \omega(Z,X)$, while the fourth produces $-\omega( -(Yf)X,Z) = (Yf)\omega(X,Z)$. These are negatives of each other because of the antisymmetry of $\omega$.