Calculating Commutator of Differential Angular Momentum
Your problem with the sign stems from two things: Firstly, you use a wrong equation for the commutator $[\hat{L}_x,\hat{L}_y]$ - your prefactor is wrong by a minus sign: The correct equation (see here) is

\begin{equation}
[\hat{L}_x,\hat{L}_y] = i \hbar \hat{L}_z
\end{equation}

Secondly, your equations for the angular momentum operators are slightly wrong because you leave out the prefactor $\frac{\hbar}{i}$ in them, although you use the correct versions in your equation [1].
So, substituting the differential operator expression for $\hat{L}_z$

\begin{equation}
 \hat{L}_z = \frac{\hbar}{i} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)
\end{equation}

into the commutator equation you should get:

\begin{equation}
[\hat{L}_x,\hat{L}_y] = i \hbar \frac{\hbar}{i} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right) = \hbar^{2} \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)
\end{equation}

As for your problem with the additional $\psi$:
Everything is fine until you let equation [1] act on a wavefunction $\psi$.

\begin{equation}
[\hat{L}_x,\hat{L}_y] \psi = \left( \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x}   \right]
+ \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z}   \right] \right) \psi
\end{equation}

You seem to make the following mistake which then leads to the appearance of an additional $\psi$ in your equations:
When you let the commutator $ [\hat{A},\hat{B}]$ act on a wavefunction $\psi$ you did the following

\begin{equation}
  [\hat{A},\hat{B}] \psi = \hat{A} \psi \hat{B} \psi - \hat{B}\psi \hat{A}\psi \qquad \qquad \text{(wrong)}
\end{equation}

which is incorrect. The correct way to do this is

\begin{equation}
  [\hat{A},\hat{B}] \psi = \hat{A} \hat{B} \psi - \hat{B} \hat{A}\psi \qquad \qquad \text{(correct)}
\end{equation}

where $\hat{A} \hat{B} \psi$ means that first $\hat{B}$ acts on $\psi$ and only then $\hat{A}$ acts on the result.
So, in the case at hand this leads to 

\begin{align}
  \left[y \frac{\hbar}{i} \frac{\partial}{\partial z}, z \frac{\hbar}{i} \frac{\partial}{\partial x}   \right] \psi &amp;= \underbrace{\left(\frac{\hbar}{i}\right)^{2}}_{= \, -\hbar^{2}} \Biggl( \underbrace{ y \frac{\partial}{\partial z} \biggl( z \frac{\partial \psi}{\partial x} \biggr)}_{= \, y \frac{\partial \psi}{\partial x} + yz \frac{\partial^{2} \psi}{\partial z \partial x}} - z y \frac{\partial^{2} \psi}{\partial x \partial z} \Biggr) \\
  &amp;= -\hbar^{2} y \frac{\partial \psi}{\partial x}
\end{align}

and

\begin{align}
  \left[z \frac{\hbar}{i} \frac{\partial}{\partial y}, x \frac{\hbar}{i} \frac{\partial}{\partial z}   \right] \psi &amp;= -\hbar^{2} \Biggl( z x \frac{\partial^{2} \psi}{\partial y \partial z} - \underbrace{ x \frac{\partial}{\partial z} \biggl( z \frac{\partial \psi}{\partial y} \biggr)}_{= \, x \frac{\partial \psi}{\partial y} + xz \frac{\partial^{2} \psi}{\partial z \partial y}} \Biggr) \\
  &amp;= \hbar^{2} x \frac{\partial \psi}{\partial y}
\end{align}

If you substitute these equations into the commutator equation you get

\begin{equation}
[\hat{L}_x,\hat{L}_y] \psi = \hbar^{2} \left(x \frac{\partial \psi}{\partial y} - y \frac{\partial \psi}{\partial x} \right)
\end{equation}

which is exactly the result you should expect.