How to compute the deformation quantizations of a polynomial Poisson algebra? As a specialist told me, given a smooth affine Poisson algebra $S$ over $\mathbb{C}$, up to a choice of certain characteristic class, one can find one of the deformation quantizations of $S$, say $A$, such that  the Poisson cohomology of $S$, suitably tensored with $\mathbb{C}[[h]]$,  is isomorphic to the Hochschild cohomology of $A$. (Maybe my memory is not very good, the above is what I remember now. )
My question is, given $S$,  how to compute explicitely $A$. 
For example, if $S=\mathbb{C}[X, Y]$ with $\{X, Y\}=XY$, how to compute the "right" deformation quantization? I guess that 
   $A=\mathbb{C}[[h]]<X, Y>/(XY-(h+1)YX)$ 
What about polynomial Poisson algebras in general? 
 A: Not a complete answer but some observations. First it might be necessary not to take formal power series but formal Laurent series in order to get a reasonable behaviour of the Hochschild cohomology. The reason is that say in the symplectic case where the relation would be $[X, Y] = \hbar 1$ you would like all derivations to be inner (first Hochschild cohomology zero). What you can prove is that the derivations are "quasi-inner" in the sense that every derivation is of the form $D = \frac{1}{\hbar} \mathrm{ad}(a)$ with some algebra element $a$. So if working over the formal power series, such derivations are "outer" since the hypothetical element $\frac{a}{\hbar}$ is not part of your algebra. Then the Hochschild cohomology has weird contributions in zeroth order of $\hbar$ and is trivial in higher order (you can check this rather easily, say for the first Hochschild cohomology)
That being said, I am aware of some results in the symplectic and smooth case e.g. by Weinstein and Xu (preprint around 1999?) where they compute the Hochschild cohomology to be essentially the deRham cohomology of the underlying manifold. You should be able to transfer these results to polynomial algebras (which have a flat space as underlying manifold, thus trivial deRham cohomology). 
And of course, in the symplectic case the deRham cohomology is the Poisson cohomology...
