"Natural" proof of $P\left(\frac{d}{dx}\right)\bigl(e^{xy}Q(x)\bigr)=Q\left(\frac{d}{dy}\right)\bigl(e^{xy}P(y)\bigr)$.

Here I post a proof based on the comment by KCd since he has not done it himself (yet): $$ P\left(\frac{\partial}{\partial x}\right)\bigl(e^{xy}Q(x)\bigr) = P\left(\frac{\partial}{\partial x}\right)Q\left(\frac{\partial}{\partial y}\right)e^{xy} = Q\left(\frac{\partial}{\partial y}\right)P\left(\frac{\partial}{\partial x}\right)e^{xy} = Q\left(\frac{\partial}{\partial y}\right)\bigl(e^{xy}P(y)\bigr). $$