How do I find the function derivative $(\delta/\delta \phi) (\partial_\mu \phi)$? The question is simple: How do I find the function derivative of $$(\delta/\delta \phi(x)) (\partial_\mu \phi(x))~?$$ As far as I can tell, I cannot use any of the standard computational rules for the functional derivative.
 A: The expression $\frac{\delta \partial_\mu\phi(y)}{\delta \phi(y)}$ is mathematically meaningless.
By definition, given a functional $F$  associating reals (or, more generally, complex numbers)  $F[\phi]$ to smooth functions $\phi$, we say that the distribution  $\frac{\delta F}{\delta \phi(x)}$ is the functional derivative of $F$, if 
$$\frac{d}{d\alpha}|_{\alpha=0} F[\phi + \alpha f] = \int \frac{\delta F}{\delta \phi(x)} f(x) dx$$
for every compactly-supported smooth function $f$.
In the considered case, one has to compute the functional derivative of the functional $F$ associating $\partial_\mu \phi(y)$ to $\phi$, i.e., 
$$\partial_\mu \phi(y) := \int \partial_\mu \phi(x) \delta(x-y) dx\:.$$
We have
$$\frac{d}{d\alpha}|_{\alpha=0} F[\phi + \alpha f] =
\frac{d}{d\alpha}|_{\alpha=0}  \int  \partial_\mu (\phi(x)+ \alpha f(x)) \delta(x-y) dx =  \int  \partial_\mu f(x) \delta(x-y) dx$$ $$= -\int   f(x) \partial_\mu\delta(x-y) dx\:.$$
We conclude that
$$\frac{\delta \partial_\mu\phi(y)}{\delta \phi(x)} = - \partial^{(x)}_\mu\delta(x-y) =  \partial^{(y)}_\mu\delta(x-y)\:. $$
So $\frac{\delta }{\delta \phi(x)}$  and $\partial^{(y)}_\mu$ commute as said by @AccidentalFourierTransform. 
In summary, $\frac{\delta \partial_\mu\phi(y)}{\delta \phi(y)}$ is not defined because the value at a fixed point of a non-regular distribution has no meaning.
A: *

*As correctly pointed out in the answer by Valter Moretti, it is mathematically ill-defined to apply (the traditional definition of) the functional/variational derivative (FD)
$$ \frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}  \tag{1}$$
to the same spacetime point.

*However, it is very common to introduce a 'same-spacetime' FD as
$$ \frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}~:=~ 
\frac{\partial{\cal L}(x) }{\partial\phi^{\alpha} (x)} - d_{\mu} \left(\frac{\partial{\cal L}(x) }{\partial\partial_{\mu}\phi^{\alpha} (x)} \right)+\ldots. \tag{2} $$
which obscures/betrays its variational origin, but is often used for notational convenience.
(The ellipsis $\ldots$ in eq. (2) denotes possible contributions from higher-order spacetime derivatives.) See e.g. this, this, & this Phys.SE posts. 
If we interpret OP's expression via eq. (2), then OP's Lagrangian density ${\cal L}=\partial_{\mu} \phi$ is a total space-time derivative, so that OP's 'same-spacetime' FD vanishes, cf. e.g. this Phys.SE post.
