Computation of the Wirtinger derivative of a product Let's have a function $f = (A/2)\phi\bar{\phi}$, where $\phi=\phi(z)$ is a complex-valued scalar field. I need to obtain $df/d\phi$. If I treat the real and imaginary parts of $\phi$ as independent variables, I can write $df/d\phi^R=A\phi^R$ and $df/d\phi^I=\phi^I$, where $R$ and $I$ stand for real and imaginary parts, respectively. The derivative above is then obtained as $df/d\phi=df/d\phi^R+i(df/d\phi^I)$, i.e. it is $df/d\phi=A\phi$. Is this not correct?
 A: If you are taking derivative with respect to $\phi$, then $\phi$ is being considered as a variable, not as a function of $z$. Assuming $A$ is constant, we have 
$$\frac{\partial}{\partial \phi} (A/2)\phi\bar \phi = (A/2)\bar \phi \tag1$$
and
$$\frac{\partial}{\partial \bar \phi} (A/2)\phi\bar \phi = (A/2)  \phi \tag2$$
The reason is that the Wirtinger derivatives   obey the product rule and  satisfy 
$$\frac{\partial}{\partial \phi} \bar \phi =0,\quad \frac{\partial}{\partial \bar\phi}  \phi =0\tag3$$ This is why we can treat $\bar \phi$ as a constant when taking the derivative with respect to $\phi$, and vice versa. 
I don't think "$df/d\phi=df/d\phi^R+i(df/d\phi^I)$" is right. The definition I know is
$$\frac{\partial}{\partial \phi} = \frac12\left(\frac{\partial}{\partial \phi^R}-i\frac{\partial}{\partial \phi^I}\right)\tag4$$
A: I assume $A$ is a constant in your context. If this is true then the whole idea of the Wirtinger calculus is that we can treat $z$ and $\bar{z}$ as independent variables. This is briefly captured by the heuristics given below (the last half of my answer justifies these as mere short-hands for a particular complex-linear combination of real partial derivatives)
$$ \frac{\partial z}{\partial \bar{z}}=0 \qquad \text{and} \qquad \frac{\partial \bar{z}}{\partial z}=0 $$
and proceed normally. In particular, if $B$ is a complex constant and $g(z,\bar{z}) = Bz\bar{z}$ then
$$ \frac{\partial g}{\partial z} = \frac{\partial }{\partial z}Bz\bar{z} = B\biggl(\frac{\partial z}{\partial z}\bar{z}+z\frac{\partial \bar{z} }{\partial z}\biggr) = B\bar{z} $$
likewise,
$$ \frac{\partial g}{\partial \bar{z}} = \frac{\partial }{\partial \bar{z}}Bz\bar{z} = B\biggl(\frac{\partial z}{\partial \bar{z}}\bar{z}+z\frac{\partial \bar{z} }{\partial \bar{z}}\biggr) = Bz $$
Now, in your context, I assume classical field theory, you consider functional derivatives with respect to complex-valued scalar fields $\phi$ and $\bar{\phi}$ and the structure is just the same. As I explain below, the derivatives with respect to $\phi$ and $\bar{\phi}$ are really just a slick notation for particular complex-linear combinations of the derivatives with respect to the real and imaginary components of $\phi$. In short, you should derive:
$$ \frac{\partial f}{\partial \phi} = \frac{A\bar{\phi}}{2} \qquad \text{and} \qquad \frac{\partial f}{\partial \bar{\phi}} = \frac{A\phi}{2} $$

The fundamental heuristics can be "derived" from realizing $\partial/\partial z$ and $\partial/\partial \bar{z}$ in terms of formal chain rules. Note $z = x+iy$ and $\bar{z} = x-iy$ invert to yield $x = \frac{1}{2}(z + \bar{z})$ and $y = \frac{1}{2i}(z-\bar{z})$
from which the following calculations seem plausible:
$$ \frac{\partial }{\partial z} = \frac{\partial x}{\partial z}\frac{\partial }{\partial x } +\frac{\partial y}{\partial z}\frac{\partial }{\partial y} = \frac{1}{2} \biggl( \frac{\partial }{\partial x }-i\frac{\partial }{\partial y } \biggr)$$
and 
$$ \frac{\partial }{\partial \bar{z}} = \frac{\partial x}{\partial \bar{z}}\frac{\partial }{\partial x } +\frac{\partial y}{\partial \bar{z}}\frac{\partial }{\partial y} = \frac{1}{2} \biggl( \frac{\partial }{\partial x }+i\frac{\partial }{\partial y } \biggr)$$
Notice that the condition $\frac{\partial f}{\partial \bar{z}}=0$ implies $f = f(z)$ which means that the function is holomorphic and one can easily verify that $f=u+iv$ will necessarily satisfy $u_x=v_y$ and $u_y=-v_x$ as a consequence of $\frac{\partial f}{\partial \bar{z}}=0$. Of course, in this context, this is not assumed for all functions and it is meaningful to write derivatives with respect to $z$ or $\bar{z}$. These should be understood as short-hands for the complex-linear combinations of real partial derivatives below:
$$ \frac{\partial f}{\partial z}  = \frac{1}{2} \biggl( \frac{\partial f}{\partial x }-i\frac{\partial f}{\partial y } \biggr) \qquad 
\text{and} \qquad 
 \frac{\partial f}{\partial \bar{z}} =  \frac{1}{2} \biggl( \frac{\partial f}{\partial x }+i\frac{\partial f}{\partial y } \biggr)$$
where if $f$ is presented at $f(z, \bar{z})$ it is understood to replace $z$ with $x+iy$ and $\bar{z} = x-iy$. In view of these definitions I can justify the heuristics as follows:
$$ \frac{\partial z}{\partial z}  = \frac{1}{2} \biggl( \frac{\partial (x+iy)}{\partial x }-i\frac{\partial (x+iy)}{\partial y } \biggr)  = \frac{1}{2}-i^2\frac{1}{2}=1.\qquad 
\text{and} \qquad 
 \frac{\partial z}{\partial \bar{z}} =  \frac{1}{2} \biggl( \frac{\partial (x+iy)}{\partial x }+i\frac{\partial (x+iy)}{\partial y } \biggr) = \frac{1}{2}+i^2\frac{1}{2}=0$$
and the proofs of $\frac{\partial \bar{z}}{\partial z}=0$ and $\frac{\partial \bar{z}}{\partial \bar{z}}=1$ are very similar.
