Show that $[\hat L_x,\hat L_y]=i\hbar \hat L_z$ The angular momentum components in Cartesians are
$$\hat L_x=\hat y\hat p_z-\hat z\hat p_y$$
$$\hat L_y=\hat z\hat p_x-\hat x\hat p_z$$
$$\hat L_z=\hat x\hat p_y-\hat y\hat p_x$$
Starting from
$[\hat L_x,\hat L_y]=\hat L_x \hat L_y-\hat L_y \hat L_x$
$\begin{align}\hat L_x\hat L_y &=(\hat y\hat p_z-\hat z\hat p_y)(\hat z \hat p_x-\hat x\hat p_z)\\&=\hat y\hat p_z\hat z\hat p_x-\hat y\hat p_z\hat x\hat p_z-\hat z\hat p_y\hat z\hat p_x+\hat z\hat p_y\hat x\hat p_z\end{align}\tag{1}$
$\begin{align}\hat L_y\hat L_x&=(\hat z \hat p_x-\hat x\hat p_z)(\hat y\hat p_z-\hat z\hat p_y)\\&=\hat z\hat p_x\hat y\hat p_z-\hat z\hat p_x\hat z\hat p_y-\hat x\hat p_z\hat y\hat p_z+\hat x\hat p_z\hat z\hat p_y \end{align}\tag{2}$
Now, to find $[\hat L_x,\hat L_y]$ I need to subtract $(2)$ from $(1)$
$\begin{align}\hat L_x \hat L_y-\hat L_y \hat L_x&=\hat y\hat p_z\hat z\hat p_x-\hat y\hat p_z\hat x\hat p_z-\hat z\hat p_y\hat z\hat p_x+\hat z\hat p_y\hat x\hat p_z\tag{3}\\&-\hat z\hat p_x\hat y\hat p_z+\hat z\hat p_x\hat z\hat p_y+\hat x\hat p_z\hat y\hat p_z-\hat x\hat p_z\hat z\hat p_y\end{align}$
Now, in order to show the relation in the title of this question, I need to commute some of the $\hat r_i$, and $\hat p_j$, subject to, 
$$[\hat r_i,\hat p_j]=i\hbar \delta_{ij}$$
where, $\hat r_i \in \{\hat x,\hat y,\hat z\}$ and $\hat p_j \in \{\hat p_x,\hat p_y,\hat p_z\}$
But, I can only commute those which have a vanishing commutator.
The 1st and 5th terms in $(3)$ are
$\hat y\hat p_z\hat z\hat p_x$ and $-\hat z\hat p_x\hat y\hat p_z$ respectively.
I know that I can swap the order of the $\hat z$ and $\hat p_x$ in the 1st term as the commutator is zero. But I cannot commute the $\hat p_z$ and $\hat z$ as the commutator is non-zero. So the first term is $\hat y\color{red}{\hat p_z\hat p_x}\hat z$. Now, I want to get the $\hat p_z$ and $\hat z$ at the end of that term; $\hat y\hat p_x\hat p_z\hat z$ 
The problem is that the $\color{red}{\hat p_x}$ and $\color{red}{\hat p_z}$ do not commute.
Put simply, I need the bracketed part $\hat y\color{red}{(}\hat p_z\hat z\color{red}{)}\hat p_x$ at the end to make any progress with this derivation. 
But how can I commute $\color{red}{(}\hat p_z\hat z\color{red}{)}$ past $\hat p_x$ if their commutator is non-zero (I know this as I have calculated it)?

From my notes it just says:

$$\begin{align}\hat L_x \hat L_y-\hat L_y \hat L_x&=\hat y\hat p_x\hat p_z\hat z-\hat y\hat x\hat p_z^2-\hat p_x\hat p_y\hat z^2+\hat x\hat p_y\hat z\hat p_z\\&-\hat y\hat p_x\hat z\hat p_z+\hat y\hat x\hat p_z^2+\hat p_x\hat p_y\hat z^2-\hat x\hat p_y\hat p_z\hat z\end{align}$$
$$\begin{align}[\hat L_x ,\hat L_y]&=\hat y\hat p_x[\hat p_z,\hat z]+\hat x\hat p_y[\hat z, \hat p_z]\\&=-i\hbar\hat y\hat p_y+i\hbar\hat x\hat p_y\\&=i\hbar \hat L_z\end{align}$$


Since those operators don't commute; How was the author of this able to reach the result above?

Edit:
According to comment $[\hat p_x,\hat p_z]=0$
But when calculating,
$$\begin{align}[\hat p_x,\hat p_z]&=-i\hbar \frac{\partial}{\partial x}\left(-i\hbar \frac{\partial}{\partial z}\right)-\left(-i\hbar \frac{\partial}{\partial z}\left(-i\hbar \frac{\partial}{\partial x}\right)\right)\\&=-\hbar^2\frac{\partial}{\partial x}\frac{\partial}{\partial z}+\hbar^2\frac{\partial}{\partial z}\frac{\partial}{\partial x}\ne 0\end{align}$$ 
Unless $$\frac{\partial}{\partial x}\frac{\partial}{\partial z}=\frac{\partial}{\partial z}\frac{\partial}{\partial x}$$
So put simply, I thought that the derivative operators must be in the same order else the commutator will be non-zero. 
Why doesn't it matter if the operators are not in the same order in both terms?
 A: I'll only do half the details so you can learn by doing the other half.
Bear in mind $r$s commute with $r$s and $p$s with $p$s, and you occasionally have what I hope are typos, such as $-i\hbar\hat{y}\hat{p}_y+i\hbar\hat{x}\hat{p}_y$.
If we accept your formula $L_i=\epsilon_{ijk}r_jp_k$, we can make progress with$$[AB,\,CD]=A[B,\,C]D+[A,\,C]BD+CA[B,\,D]+C[A,\,D]B$$(9. here). Here's a brief summary of what happens when we use the known phase space commutators, plus famous Levi-CIvita identities:$$\begin{align}[L_i,\,L_j]&=\epsilon_{ilm}\epsilon_{jno}[r_lp_m,\,r_np_o]\\&=i\hbar(\epsilon_{ilm}\epsilon_{jnl}r_np_m-\epsilon_{ilm}\epsilon_{jmo}r_lp_o)\\&=i\hbar(r_ip_j-r_jp_i)\\&=i\hbar\epsilon_{ijk}L_k.\end{align}$$While this is the desired result, this entire exercise has one subtle failing: since we want angular momentum to be Hermitian, we should have started from $L_i=\frac12\epsilon_{ijk}(r_jp_k-p_jr_k)$. But you can repeat the other techniques to derived $[L_i,\,L_j]=i\hbar\epsilon_{ijk}L_k$.
A: The trick is to rewrite $p_i x_i$ as $[p_i, x_i] + x_i p_i$.
First we expand $[L_x, L_y]$:
$$
[L_x, L_y]
= [y p_z - z p_y, z p_x - x p_z]
= [y p_z, z p_x] - [z p_y, z p_x] - [y p_z, x p_z] + [z p_y, x p_z]
$$
Then we expand each of the above commutators:
$$\begin{align}
[y p_z, z p_x]
&= y p_z \, z p_x - z p_x \, y p_z
= y ([p_z, z] + z p_z) p_x - y z p_x p_z
= y (-i\hbar + z p_z) p_x - y z p_x p_z
= -i\hbar y p_x \\
[z p_y, z p_x]
&= z p_y \, z p_x - z p_x \, z p_y
= z^2 p_x p_y - z^2 p_x p_y
= 0 \\
[y p_z, x p_z]
&= y p_z \, x p_z - x p_z \, y p_z
= xy p_z^2 - xy p_z^2
= 0 \\
[z p_y, x p_z]
&= z p_y \, x p_z - x p_z \, z p_y
= xz p_y p_z - x ([p_z, z] + z p_z) p_y
= xz p_y p_z - x (-i\hbar + z p_z) p_y
= i\hbar x p_y
\end{align}$$
Thus,
$$
[L_x, L_y]
= -i\hbar y p_x - 0 - 0 + i\hbar x p_y
= i\hbar (x p_y - y p_x)
= i\hbar L_z
.
$$
