How to apply the definition of complex conjugate to a partial derivative In my book on complex variables and applications, the concept of a complex conjugate is simply stated as a reflection about the real axis in the complex plane. Mathematically speaking, the complex conjugate of $x+iy$ is $x-iy$. They then go on to describe some properties of the conjugate and provide some rather "simple" examples. My book is likely an introduction to the subject, for which I'm not complaining, but I've now encountered a more difficult problem and lack the confidence to move forward.
I'm working on a problem, beyond the scope of examples given in the book, where an expression is given as 
$\displaystyle \frac{1}{\frac{\partial f(z)}{\partial z}} \frac{f(z)}{z}$
in which $z$ is complex. Can someone help step me through how to find the complex conjugate of the expression? Can you instruct me such that I can carry the knowledge and apply it to other practical/applied problems. In other words, instead of just giving the answer, tell me how/why you obtain what you did. For example, if $f(z)$ is real, I could possibly get one answer, whereas if $f(z)$ is complex, I'd possibly get a different one. I don't know, but I'd like to know how to determine the conjugate in either case.
EDIT: So here's my "stab" at it following the rules presented in my book
$\displaystyle \overline{\frac{1}{\frac{\partial f(z)}{\partial z}} \frac{f(z)}{z}} =\overline{xy} =\overline{x}\:\overline{y}$
where
$\displaystyle x = \frac{1}{\frac{\partial f(z)}{\partial z}} \quad,\quad y = \frac{f(z)}{z}$
Looking first at the y-term, the conjugate of a rational expression is equal to a rational of the conjugates which gives
$\displaystyle \overline{y} = \frac{\overline{f(z)}}{\overline{z}}$
Applying the same logic to the x-term gives
$\displaystyle \overline{x} = \frac{\overline{1}}{\overline{\frac{\partial f(z)}{\partial z}}}$
The complex conjugate of 1 is just itself, so bringing it all together, I get
$\displaystyle \frac{1}{\overline{\frac{\partial f(z)}{\partial z}}} \frac{\overline{f(z)}}{\overline{z}}$
While the expression above may be correct, I think I can say more about the conjugate of the partial derivative term
$\displaystyle \overline{\frac{\partial f(z)}{\partial z}}$
Under what condition does the conjugate expression further simplify? Can it reduce to something like $\frac{\partial \overline{f(z)}}{\partial \overline{z}}$? Under some conditions, I think I've seen people pull the conjugate into the function as well i.e. $\overline{f(z)} = f(\overline{z})$. Basically, I want to know if I can say more about $\overline{\frac{\partial f(z)}{\partial z}}$ and what conditions must be satisfied to perform such manipulations.
 A: Let $f(z)=u(x,y)+iv(x,y)$.  Then, note that since $z=x+iy$ and $\bar z=x-iy$, we have $x=\frac12(z+\bar z)$ and $y=\frac1{i2}(z-\bar z)$
$$\begin{align}
\frac{\partial f(z)}{\partial z}&=\frac12\frac{\partial f(z)}{\partial x}-\frac i2\frac{\partial f(z)}{\partial y}\\\\
&=\frac12\left(\frac{\partial u(x,y)}{\partial x}+\frac{\partial v(x,y)}{\partial y}\right)+\frac i2\left(-\frac{\partial u(x,y)}{\partial y}+\frac{\partial v(x,y)}{\partial x}\right)\tag 1
\end{align}$$
Taking the complex conjugate of $(1)$ yields
$$\begin{align}
\overline{\frac{\partial f(z)}{\partial z}}&=\frac12\left(\frac{\partial u(x,y)}{\partial x}+\frac{\partial v(x,y)}{\partial y}\right)-\frac i2\left(-\frac{\partial u(x,y)}{\partial y}+\frac{\partial v(x,y)}{\partial x}\right)\\\\
&=\left(\frac12\frac{\partial }{\partial x}+\frac i2 \frac{\partial }{\partial y}\right)(u(x,y)-iv(x,y)) \\\\
&=\frac{\partial \overline{f(z)}}{\partial \bar z}
\end{align}$$
So, we see that if $\frac{\partial f(z)}{\partial z}$ exists, then its complex conjugate is given by 
$$\overline{\frac{\partial f(z)}{\partial z}}=\frac{\partial \overline{f(z)}}{\partial \bar z}$$
It is not true in general that $\overline{f(z)}$ is equal to $f(\bar z)$.  If $u(x,y)-iv(x,y)=u(x,-y)+iv(x,-y)$, then we can write
$$\overline{\frac{\partial f(z)}{\partial z}}={\frac{\partial f(\bar z)}{\partial \bar z}}$$
