To compute complex derivatives Leibniz style Here's a semi-serious question. 
Here we have introduced the differential forms 
$$\tag{1}
\begin{array}{cc}
dz=dx+idy, & d\overline{z}=dx-idy,
\end{array}
$$
Now let us look at this formula, familiar from complex variable theory:
$$\tag{2}\begin{array}{cc}
\frac{d}{dz}=\frac{1}{2}\left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right), & \frac{d}{d\overline{z}}=\frac{1}{2}\left( \frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)
\end{array}
$$

Question. Can we recover (2) from (1) with a formal computation that treats $dx$ and $dy$ as infinitesimals? 

Inserting blindly $(1)$ into the left hand sides of $(2)$ we get 
$$\begin{array}{cc}
\frac{d}{dz}=\frac{d}{dx+idy},&\frac{d}{d\overline{z}}=\frac{d}{dx-idy}
\end{array} $$ 
but how to move on from there? 
 A: You shouldn't view $\dfrac{d}{dx}$ as the quotient of $d$ (whatever that would be) and $dx$ but rather as the dual of $dx$ in the sense that
$$
df = \frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy.
$$
Similarly,
$$
df = \frac{\partial f}{\partial z}\,dz + \frac{\partial f}{\partial \bar z}\,d\bar z.
$$
Here, $df$ is the exterior derivative of $f$.
Equating the two and using $dz = dx + i\,dy$ and $d\bar z = dx - i\,dy$ you get
\begin{align}
\frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy &=
\frac{\partial f}{\partial z}\,dz + \frac{\partial f}{\partial \bar z}\,d\bar z \\
&= \frac{\partial f}{\partial z}\,(dx + i\,dy) + \frac{\partial f}{\partial \bar z}\,(dx -i\,dy) \\
&= \left( \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \bar z} \right)\,dx 
+ i\left( \frac{\partial f}{\partial z} - \frac{\partial f}{\partial \bar z} \right)\,dy.
\end{align}
I.e.
$$
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \bar z}
$$
and
$$
\frac1i \frac{\partial f}{\partial y} = \frac{\partial f}{\partial z} - \frac{\partial f}{\partial \bar z}.
$$
Adding and subtracting these two equations finally gives you
\begin{align}
\frac{\partial f}{\partial z} &= 
\frac12 \left( \frac{\partial f}{\partial x} + \frac1i \frac{\partial f}{\partial y}\right)
= \frac12 \left( \frac{\partial f}{\partial x} -i \frac{\partial f}{\partial y}\right) \\
\frac{\partial f}{\partial \bar z} &= 
\frac12 \left( \frac{\partial f}{\partial x} - \frac1i \frac{\partial f}{\partial y}\right)
= \frac12 \left( \frac{\partial f}{\partial x} +i \frac{\partial f}{\partial y}\right) \\
\end{align}
as expected.
