About $f(z)dz$(closed differential form)

Question

I encounter a question as I try to show that $f(z)dz$ is a closed differential form, for any holomorphism function $f$, $$d(f(z)dz)=\partial_{z}fdz \wedge dz+\partial_{\bar{z}}fd\bar{z} \wedge dz=0.$$

But I calculate that $d\bar{z} \wedge dz= (dx-idy)\wedge (dx+idy)= 2idx\wedge dy$. How to get the result?

Answer 1

If $f$ is holomorphic then $f_{\bar z}= 0$. (Actually, its "if and only" if, assuming $f$ is differentiable in some sense).