= Solution
A <Poisson manifold> is a smooth <manifold> $P$ with a bilinear bracket on $C^\infty(P)$ that is antisymmetric, satisfies the <Jacobi identity> and obeys $\{f,gh\}=\{f,g\}h+g\{f,h\}$. Equivalently it has a <bivector> $\Pi$ defining $\{f,g\}=\Pi(df,dg)$ with vanishing Schouten bracket $[\Pi,\Pi]$; nondegeneracy is not required.
For a <symplectic manifold>, define the <Hamiltonian vector field> by $\iota_{X_f}\omega=df$ and put $\{f,g\}=\omega(X_f,X_g)$. Nondegeneracy gives a unique $X_f$, antisymmetry is inherited from $\omega$, and $d(gh)=g\,dh+h\,dg$ proves the <Leibniz rule>. Moreover $\mathcal L_{X_f}\omega=d\iota_{X_f}\omega+\iota_{X_f}d\omega=0$ by <Cartan's magic formula>. Since $X_fg=-\{f,g\}$,
$$
\iota_{[X_f,X_g]}\omega=\mathcal L_{X_f}(\iota_{X_g}\omega)=d(X_fg)=-d\{f,g\},
\qquad [X_f,X_g]=-X_{\{f,g\}}.
$$
For completeness, evaluate $d\omega=0$ on $X_f,X_g,X_h$. Its three derivative terms sum to minus the cyclic sum of $\{f,\{g,h\}\}$, and its three bracket terms sum to minus the same cyclic sum, using the displayed identity. Hence that sum is zero. This proves the <Jacobi identity for the Poisson bracket>, not merely the Jacobi identity modulo constants. The inverse of a closed nondegenerate two-form therefore defines a <Poisson manifold>.
The <Symplectic Darboux theorem> says that near every point of a $2n$-dimensional <symplectic manifold> there are coordinates $(q^1,\ldots,q^n,p_1,\ldots,p_n)$ with
$$
\boxed{\omega=\sum_{i=1}^n dq^i\wedge dp_i,\qquad
\{f,g\}=\sum_i\left(\frac{\partial f}{\partial q^i}\frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q^i}\right).}
$$
Thus every symplectic structure has the same local normal form, despite possible global topological differences.
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