= Solution
A <Poisson manifold> is a smooth <manifold> with an antisymmetric contravariant <tensor field> $\pi=\tfrac12\pi^{ij}\partial_i\wedge\partial_j$ such that the bracket
$$
\{f,g\}=\pi^{ij}(\partial_i f)(\partial_j g)
$$
satisfies the <Jacobi identity>. Bilinearity, antisymmetry and the <Leibniz rule> follow directly from this formula. The <Poisson bivector> is allowed to be degenerate.
Apply the <Jacobi identity> to coordinate functions $x^i,x^j,x^k$. Necessity gives the <coordinate Jacobi condition for a Poisson bivector>:
$$
\boxed{J^{ijk}:=\pi^{i\ell}\partial_\ell\pi^{jk}+\pi^{j\ell}\partial_\ell\pi^{ki}+\pi^{k\ell}\partial_\ell\pi^{ij}=0.}
$$
For sufficiency, expand $\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}$. The terms differentiating $\pi$ are $J^{ijk}f_i g_j h_k$. All terms containing second derivatives cancel. For example, those containing $g_{j\ell}$ have coefficient $(\pi^{i\ell}\pi^{jk}+\pi^{k\ell}\pi^{ij})f_i h_k$, which is antisymmetric under $j\leftrightarrow\ell$ after contraction with $f_i h_k$, whereas $g_{j\ell}$ is symmetric. The same argument handles the second derivatives of $f$ and $h$. Thus the displayed condition is both necessary and sufficient.
A <symplectic manifold> has a smooth nondegenerate <closed differential form> $\Omega$ of degree two. Write $\Omega=\tfrac12\Omega_{ij}dx^i\wedge dx^j$ and choose the explicit inverse convention $\pi^{ik}\Omega_{kj}=\delta^i_j$. This defines a smooth antisymmetric <bivector>. Differentiating the inverse matrix gives
$$
\partial_\ell\pi^{ij}=-\pi^{ia}(\partial_\ell\Omega_{ab})\pi^{bj}.
$$
Substitution into the Jacobi coefficient, followed by relabeling summed indices, gives
$$
J^{ijk}=\pi^{ia}\pi^{jb}\pi^{kc}\big(\partial_a\Omega_{bc}+\partial_b\Omega_{ca}+\partial_c\Omega_{ab}\big)=0,
$$
since the parentheses are the components of $d\Omega$. Hence \b[every symplectic manifold carries a Poisson structure]. The opposite overall convention for the inverse bracket gives the same conclusion. Nondegeneracy and closure together are crucial; an arbitrary invertible antisymmetric tensor need not satisfy the <Jacobi identity>.
For a finite-dimensional <Lie algebra> $\mathfrak g$, its <dual space> $\mathfrak g^*$ carries the <Lie-Poisson bracket>
$$
\{f,g\}(\ell)=\ell([df_\ell,dg_\ell]),
$$
where $df_\ell,dg_\ell$ are identified with elements of $\mathfrak g$. If $[e_i,e_j]=c_{ij}{}^k e_k$ and $x_i(\ell)=\ell(e_i)$, this reads $\{x_i,x_j\}=c_{ij}{}^k x_k$. The Jacobi coefficient is
$$
\big(c_{i\ell}{}^m c_{jk}{}^\ell+c_{j\ell}{}^m c_{ki}{}^\ell+c_{k\ell}{}^m c_{ij}{}^\ell\big)x_m,
$$
which vanishes by the <Lie algebra> <Jacobi identity>.
In particular, the dual of $\mathfrak{so}(3)$ is $\mathbb R^3$ with the <rotational Lie-Poisson structure on R3>,
$$
\{x,y\}=z,\quad\{y,z\}=x,\quad\{z,x\}=y,\qquad
(\pi^{ij})=\begin{pmatrix}0&z&-y\\-z&0&x\\y&-x&0\end{pmatrix}.
$$
The <Poisson bivector> has rank two away from the origin and rank zero at the origin. The function $x^2+y^2+z^2$ is a <Casimir function>, since its bracket with every function vanishes. The centred spheres are its two-dimensional <symplectic leaves>, and the origin is a point leaf. \b[This is a Poisson manifold but not a symplectic manifold]: its bivector is not invertible, and an odd-dimensional manifold cannot have a nondegenerate alternating <bilinear form>.
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