= Solution
On the smooth-function space, the product rule gives $ABf=(xf)'=f+xf'$ while $BAf=xf'$. Hence $\boxed{[A,B]=I}$, and both maps preserve the stated smooth space.
For general operators with this <commutator> relation, use $[A,BC]=[A,B]C+B[A,C]$. Induction gives $[A,B^i]=iB^{i-1}$ for $i\geq1$. Since $Ay=0$,
$$
\boxed{A(By)=y,\qquad A(B^iy)=iB^{i-1}y\ (i\geq1),\qquad Ay=0.}
$$
Every displayed vector lies in $W$, so $W$ is invariant under $A$, as well as under $B$. Iterating the lowering formula gives $A^kB^iy=i!B^{i-k}y/(i-k)!$ for $k\leq i$, and zero for $k>i$.
Suppose a nontrivial finite <linear dependence> has highest nonzero term $c_nB^ny$. Applying $A^n$ kills all lower terms and yields $c_nn!y=0$. The field is real, $n!\ne0$, and $y\ne0$, so this contradicts $c_n\ne0$. Thus \b[the entire sequence is linearly independent]. This <identity commutator cyclic ladder> also proves that such a representation with a nonzero vector in the kernel of $A$ cannot be finite dimensional.
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