Commutator of $V(\hat{\vec{r}})$ and $\hat{L_z}$ 
Can someone explain to me why this is true?
to me I see that $$x[\hat{p_y},V(r)]- y[\hat{p_x},V(r)]$$
$$=x(\hat{p_y}V(r)- V(r)\hat{p_y} )- y (\hat{p_x}V(r)+ \hat{p_x}V(r)).$$
The only way this can equal what is in the image is if $V(r)\hat{p_y} = 0$ and $V(r)\hat{p_x}=0$ but why would that be ?
 A: As @CosmasZachos' comment already pointed out, the equation
$$\left[p_y,V(\vec{r})\right]= -i\hbar \frac{\partial V(\vec{r})}{\partial y}$$
is an operator equation. To prove such an operator equation
you need to show that the operators on the left and on the right
side yield the same result when operating on any arbitrary
wave function $\psi(\vec{r})$. So let us do this:
$$\begin{align}
\left[p_y,V(\vec{r})\right]\psi(\vec{r})
&=-i\hbar\left[\frac{\partial}{\partial y},V(\vec{r})\right]\psi(\vec{r}) \\
&=-i\hbar\left(\frac{\partial}{\partial y}V(\vec{r})
  -V(\vec{r})\frac{\partial}{\partial y}\right)\psi(\vec{r}) \\
&=-i\hbar\left(\frac{\partial}{\partial y}(V(\vec{r})\psi(\vec{r}))
  -V(\vec{r})\frac{\partial}{\partial y}\psi(\vec{r})
  \right) \\
&=-i\hbar\left(\frac{\partial V(\vec{r})}{\partial y}\psi(\vec{r})
  +V(\vec{r})\frac{\partial \psi(\vec{r})}{\partial y}
  -V(\vec{r})\frac{\partial\psi(\vec{r})}{\partial y}
  \right) \\
&=-i\hbar\frac{\partial V(\vec{r})}{\partial y}\psi(\vec{r})
\end{align}$$
Since this equation holds for every $\psi(\vec{r})$
both operators acting on $\psi(\vec{r})$ must be equal.
$$\left[p_y,V(\vec{r})\right]
=-i\hbar\frac{\partial V(\vec{r})}{\partial y}$$
