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
$$\overline{\frac{\partial f(z)}{\partial z}}={\frac{\partial f(\bar z)}{\partial \bar z}}$$
Best Answer
$-\bar{z}$ (or $-z^\ast$ if you're of that persuasion) is fine for your needs; no need for new notation.
Edit 9/29/2011:
One paper refers to the operation $-\bar{z}$ as "paraconjugation" and uses $z^\ast$ for it, but since $z^\ast$ is also often used for conjugation proper, I can't recommend the notation in good conscience. (I was hard-pressed to find other papers using the same term, also.)