Let $X$ be a non-compact complex manifold of dimension at least 2 equipped with a Kahler metric $\omega$. Take a smooth positive function $f : X \to \mathbb R$, and define a new hermitian metric on $X$ by $\tilde \omega = f \omega$. If $f$ is non-constant, then can this new metric ever be Kahler?

If $\dim_{\mathbb C} X = 1$ the new metric is automatically Kahler because of dimension. If $\dim_{\mathbb C} X \geq 2$ and if $X$ is compact the new metric is never Kahler. Indeed, we have that $d \tilde \omega = d f \wedge \omega$ is zero if and only if $df$ is zero by the hard Lefschetz theorem, so $f$ must be constant if $\tilde \omega$ is Kahler.

If $X$ is not compact, then to the best of my knowledge we do not have the hard Lefschetz theorem, but does the conclusion on metrics conformal to a Kahler metric still hold?