ra.rings and algebras - Tensor products and two-sided faithful flatness

Here's an example. Let $R = {mathbb C}[x]$ and let $S = {mathbb C}langle x,yrangle/(xy-yx-1)$, i.e. the first Weyl algebra $A_1$. Then $S$ is free as both a left and right
$R$-module, and comes equipped with the natural ($R$-bimodule) inclusion of $R$. On the
other hand, if you take $M = {mathbb C}[x]/(x) = N$, you'll get for $Motimes_R Sotimes_R M$ the zero module. Indeed, any element of $S$, i.e. any differential operator with polynomial coefficients (writing $partial = partial/partial x$ in place of $y$) can be written in the form $sum_i p_i(x) partial^i$, so any element in $Motimes_R S = {mathbb C}[x]/(x) otimes S$ is represented by an expression $sum_i c_i partial^i$ where the $c_i$ are constants, and now an induction on $k$ shows (I believe, my brain is a little fuzzy at this hour) that, for the right $R$-module structure on $S/xS$, one has $partial^k cdot x = k partial^{k-1}$. One can conclude that $Motimes_R Sotimes_R M = 0$, whereas of course $Motimes_R Mcong M$ in this example...