 Solution

On a finite-dimensional $A$-module $M$, the actions of $X$ and $Y$ are endomorphisms $P,Q$ satisfying $PQ-QP=I_M$. Taking traces and using $\mathrm{tr}(PQ)=\mathrm{tr}(QP)$ givesSince $k$ has characteristic zero, this forces ${\dim }_{k}M=0$. Thus the zero module is the only finite-dimensional module for this Weyl algebra. The obstruction is the Weyl algebra has no nonzero finite-dimensional modules.