What does $[\frac{\partial }{\partial x},y]$ look like? I refer to the first page of this paper. It says that $[\frac{\partial }{\partial x},y]=0$, where $[]$ is the Lie bracket operation.
How do we see this? Under the standard Lie bracket operation $[a,b]=ab-ba$, how would one expand $[\frac{\partial }{\partial x},y]$?
 A: The brackets are expanded exactly the way you're​ used to. So we have
$$
[\frac{\partial}{\partial x}, y]=\frac{\partial}{\partial x}y-y\frac{\partial}{\partial x}
$$
where $\frac{\partial}{\partial x}$ is the operator "partial differentiation with respect to $x$", and $y$ is the operator "multiply by the function $f(x,y)=y$", and operators are applied right-to-left. In order to do calculations, it is often advisable to apply the operator to a dummy function, $g$. We get
$$
[\frac{\partial}{\partial x}, y]g=(\frac{\partial}{\partial x}y-y\frac{\partial}{\partial x})g\\
=\frac{\partial}{\partial x}(fg)-f\frac{\partial}{\partial x}g
$$
And you can now see that this results in $0$; $f$ doesn't depend on $x$, so multiplying with it before or after the partial differentiation makes no difference.
Note that one usually writes $y$ instead of $f$. That means that the letter $y$ is used both for the function and for the operator. This is unfortunate, but it's the way physicists roll.
A: Coming from physics we generally expand Lie brackets as we see fit and seldom care whether the resulting expression is mathematically correct as written. One should not forget, that in general there's a function this operator is acting on, so:
$[\partial_x,y] = 0$ means actually $ [\partial_x,y]f(x,y)=\partial_x yf(x,y) - y\partial_x f(x,y) = 0$
