Prove $dω = 0 \iff$ {·, ·} is a Poisson bracket knowing that $\{f, g\} := X_f (g)$ and $i_{X_f} \omega = df$

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Let $\omega \in \Omega^2(M)$ be a non-degenerate 2-form. For a function $f \in C^\infty(M)$, define the vector field $X_f ∈ \mathscr X(M)$ via the relation: $$i_{X_f} \omega = df$$, where the first member is the interior multiplication and define the following binary operation on $ C^\infty(M)$ $$\{f, g\} := X_f (g)$$.

Show that

$dω = 0 \iff$ {·, ·} is a Poisson bracket.

I am new to the concept of interior multiplication and not very comfortable manipulating differentials, so I have no idea how to go about solving this. Any help is appreciated