Hamiltonian vector fields for standard Poisson structure Given a Poisson manifold $(M,\Pi)$, a vector field $X \in \mathfrak{X}(M)$ is called $\textit{Hamiltonian}$ if $\iota_X \Pi = df$ for some smooth $f$. A vector field $X$ that satisfies $\mathcal{L}_X\Pi=0$ is called $\textit{Poisson}$. 
In general, Poisson vector fields need not be Hamiltonian. However, on $(\mathbb{R}^{2n}, \sum \frac{\partial}{\partial x_i} \wedge \frac{\partial}{\partial y_i})$, every Poisson vector field is Hamiltonian. I wanted to prove this, however I can't seem to find how to begin. My first guess would be to use some analogous "Cartan magic formula", but I don't think this makes sense in the case of multi-vector fields (at least not immediately).
Any help is apprciated. 
 A: A side note: you inverted the vector field $X$ and the 1-form $df$ in the definition of Hamiltonian vector field. $X$ is Hamiltonian if it is a Poisson gradient, and some equivalent notations are $X_{f} = \{f, \cdot\} = \pi(df, \cdot) = (df)^{\sharp} = \iota_{df}\pi $. 
On a similar line, since you cannot feed vector fields to a bivector as you do with a 2-form, you cannot have a strict analogue of Cartan's magic formula for the Lie derivative of a bivector. 
Back to your question
Page 122 and around there of 

Paulette Libermann and Charles-Michel Marle. Symplectic Geometry and
  Analytical Mechanics

can be of help without invoking Poisson cohomology, that requires building Poisson calculus first (that is, defining brackets of 1-forms from usual Poisson brackets of smooth functions). 
As @studiosus said, your Poisson manifold is special in two independent ways: 
A. in $\mathbb{R}^n$ differential forms are closed iff exact (all de Rham cohomology groups are trivial); and
B. the Poisson bivector is non-degenerate everywhere (making the manifold symplectic).
These implications show how these two facts make any Poisson vector field a Hamiltonian vector field:

The arrows marked with (A) or (B) hold only if the respective condition holds; let's start climbing up this chain from the bottom. 
A vector field $X$ is called Poisson if $\mathcal{L}_X \pi = 0$.
The first difficulty we face is that a degenerate bivector $\pi$ defines a non-bijective homomorphism $\sharp$ mapping a 1-form into a vector field, so for a generic vector field there may not exist a 1-form $\alpha$ such that $\alpha^{\sharp} = X$. 
If a Poisson vector field belongs to the image of $\sharp$, so $\mathcal{L}_X \pi = 0$ and $X = \alpha^{\sharp}$ for some 1-form $\alpha$, we say that $X$ is locally Hamiltonian.
Proposition 10.5 at page 122 of Libermann's book shows that a vector field is locally Hamiltonian iff the related 1-form is c-closed, namely $d\alpha$ vanishes on hamiltonian vector fields.
Now, if B. holds, the Poisson bivector is non-degenerate, and it defines an isomorphism between vector fields and 1-forms: in particular every vector field is associated to precisely one 1-form [* footnote].
This means that, if B. holds, every Poisson vector field is locally hamiltonian, and it can be shown that every c-closed 1-form is closed (every vector can be written as the image of a Hamiltonian vector field at one point).
To summarize, if the Poisson bivector is non-degenerate, there is a 1-to-1 correspondence between Poisson vector fields and closed 1-forms.
Finally, as said in A., in $\mathbb{R}^n$ a closed 1-form is exact: $\alpha = df$ for some function $f$; so the Poisson vector field we started with, $X = \alpha^{\sharp}$, is indeed Hamiltonian: $X = (df)^{\sharp}$.

[* footnote] If B. holds the Poisson manifold is symplectic. The symplectic form $\omega$ can be defined from the bivector since $\sharp^{-1}$ now makes sense; the non-degeneracy of $\pi$ grants the non-degeneracy of $\omega$; and interestingly the Jacobi property of the Poisson structure, or equivalently the vanishing Schouten brackets $[\pi,\pi]_S = 0$, assures that $d \omega = 0$. 
