Differentiating D'Alembert operator It has been a while since I did field theory.
Euler-Lagrange equation 
$$\partial_\mu \frac{\partial L}{\partial (\partial _\mu \phi)} - \frac{\partial L}{\partial \phi} = 0$$
If I have 
$$L = \phi \Box \phi - m^2 \phi^2,$$ 
do we just get
$$\Box \phi - 2 m^2 \phi = 0$$ 
Because we don't differentiate the D'Alembert operator?
 A: In the interest of completeness, a Lagrangian (density) which includes second derivatives generally presents significant technical difficulties.  Note that $L$ is a function, not a functional, which means that it takes $\phi$,$\partial \phi$, and $\partial \partial \phi$ as independent inputs (in other words, $\mathcal L$ does not "know how" to take derivatives).
As a result, varying the action leads to
$$\delta S = \int d^4x \left(\frac{\partial \mathcal{L}}{\partial \phi} - \partial_\mu \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} + \partial_\mu\partial_\nu \frac{\partial \mathcal{L}}{\partial(\partial_\mu\partial_\nu \phi)}\right)\delta \phi$$
$$+ \oint \hat n_\mu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} - \partial_\nu\frac{\partial \mathcal{L}}{\partial(\partial_\mu\partial_\nu \phi)}\right) \delta \phi \ dS+\oint \hat n_\mu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu\partial_\nu\phi)} \partial_\nu(\delta\phi)\right)dS$$
The usual demand that $\delta\phi=0$ on the boundaries is generically insufficient to render this a well-defined variational problem, because the last surface integral would not generally go away.  For that, we would need to also demand that $\partial_\nu(\delta \phi)=0$ on the boundaries, which means that we would need to provide boundary conditions on the field $\phi$ and all of its derivatives $\partial_\nu \phi$.
This is where the trouble comes in.  If we fix $\phi$ and its derivatives on the boundary, we obtain the Euler-Lagrange equations
$$\frac{\partial\mathcal{L}}{\partial \phi} - \partial_\mu \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} + \partial_\mu\partial_\nu \frac{\partial \mathcal{L}}{\partial(\partial_\mu\partial_\nu \phi)} = 0$$
For your Lagrangian,
$$\frac{\partial \mathcal{L}}{\partial \phi} =\square \phi - 2m^2\phi = \eta^{\mu\nu}\partial_\mu\partial_\nu \phi -2 m^2\phi$$
$$\frac{\partial \mathcal{L}}{\partial (\partial_\mu\phi)}=0$$
$$\frac{\partial \mathcal{L}}{\partial (\partial_\mu\partial_\nu\phi)} = \eta^{\mu\nu}\phi$$
And so we have that (cancelling a factor of $2$)
$$\square \phi -m^2\phi^2=0$$
The trouble with this is that it is a second order PDE where we've fixed the field and its derivatives on the boundaries of the spacetime region under consideration.  That's too many boundary conditions, and does not generally yield a consistent solution except under miraculously fortuitous circumstances.
Incidentally, this is precisely what happens in general relativity.  The solution is to add a surface term to the original action which cancels out the problematic surface integral, freeing us from the need to set $\partial_\nu(\delta \phi)$ to zero.  In this case, the appropriate action is
$$S = \int d^4x \mathcal{L} - \oint \hat n_\mu \eta^{\mu\nu}\partial_\nu \phi \ dS$$
while in general relativity, the appropriate correction is called the Gibbons-Hawking-York boundary term.
A: Since the Lagrangian contains second derivative, you will need to use
$$ \frac{\partial L}{\partial \phi}-\partial_\mu \frac{\partial L}{\partial (\partial _\mu \phi)} +\partial_{\mu}\partial_\nu \frac{\partial L}{\partial (\partial _\mu\partial_\nu \phi)}  = 0$$
which yields equation of motion
$$\square\phi-m^2\phi=0$$
A: Focus on the first term of the Lagrangian
$$ \phi \Box \phi = \phi \partial^\nu\partial_\nu \phi $$
and act with the Euler-Lagrange operator $$\partial_\mu \frac{\partial}{\partial (\partial _\mu \phi)} - \frac{\partial}{\partial \phi}$$
Evaluating the second part readily gives
$$ - \frac{\partial}{\partial \phi}\phi \partial^\nu\partial_\nu \phi = -\partial^\nu\partial_\nu \phi $$ which is already the result you correctly obtained. The first part evaluates to zero precisely because the Lagrangian does not contain a first derivative in $\phi$ - only zeroth and second.
I assume that is what you mean by

Because we don't differentiate the D'Alembert operator  

Indeed,
$$\partial_\mu \frac{\partial}{\partial (\partial _\mu \phi)} \phi \partial^\nu\partial_\nu \phi = \partial_\mu \phi\partial^\nu\delta_{\mu\nu} = 0$$
One could obtain the same result by integrating the Langrangian by parts
$$ L = -\partial^\nu\phi \, \partial_\nu \phi -m^2\phi^2= -\eta^{\kappa\nu} \partial_\kappa\phi\,\partial_\nu\phi -m^2\phi^2$$
