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You don't have to use hard Lefschetz to conclude $df=0$ from $\omega\wedge df=0$.

This is a linear algebra fact, valid pointwise : if $\alpha \in T_x^*X$ satisfies $\omega_x \wedge \alpha=0$, then $\alpha=0$ (of course, assuming $\dim_R X \geq 4$.

The short argument is that, $\omega_x^{n-1}\wedge : T^*_x X\to \bigwedge^{2n-1} T^*_x X$ is an isomorphism ("pointwise not so hard Lefschetz", so to speak).

This said, as in Francesco's answer, you can have non proportional conformal riemannian metrics that are Kähler with respect to different complex structures, so that the corresponding 2-forms are no longer (pointwise) proportional.

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You don't have to use hard Lefschetz to conclude $df=0$ from $\omega\wedge df=0$.

This is a linear algebra fact, valid pointwise : if $\alpha \in T_x^*X$ satisfies $\omega_x \wedge \alpha=0$, then $\alpha=0$ (of course, assuming $\dim_R X \geq 4$.

The short argument is that, $\omega_x^{n-1}\wedge : T^*_x X\to \bigwedge^{2n-1} T^*_x X$ is an isomorphism ("pointwise not so hard Lefschetz", so to speak).