Let us think formally of $z$ and $z^*$ as independent variables for the plane, like $x$ and $y$. Then we have that $$ \frac{\partial f}{\partial z*}= \frac{\partial f}{\partial(x-iy)}= \frac{\partial f}{\partial x}+i \frac{\partial f}{\partial y}. $$
Can you explain how to derive the above equation for complex numbered functions?
We have $$ x = \frac{1}{2}(z+z^*), \quad y = \frac{1}{2i}(z-z^*), $$ so $$ \frac{\partial f}{\partial z^*} = \frac{\partial x}{\partial z^*} \frac{\partial f}{\partial x} + \frac{\partial y}{\partial z^*} \frac{\partial f}{\partial y} = \frac{1}{2} \frac{\partial f}{\partial x} - \frac{1}{2i}\frac{\partial f}{\partial y} = \frac{1}{2} \left( \frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y} \right) $$ (and in particular, the formula in the question is missing a factor of $1/2$).
(note: the formula in the OP is wrong)
Differentials tend to be better behaved, especially when multiple related variables are involved. The crucial equations are
$$ z = x + \mathbf{i} y\qquad \qquad z^* = x - \mathbf{i} y$$
which differentiates to
$$ \mathrm{d}z = \mathrm{d} x + \mathbf{i} \mathrm{d} y \qquad \qquad \mathrm{d}z^* = \mathrm{d} x - \mathbf{i} \mathrm{d} y $$
which we can solve to get
$$ \mathrm{d}x = \frac{\mathrm{d}z + \mathrm{d}z^*}{2} \qquad \qquad \mathrm{d}y = \frac{\mathrm{d}z - \mathrm{d}z^*}{2 \mathbf{i}} $$
which we can apply if we need to switch from the $\mathrm{d}x, \mathrm{d}y$ basis to the $\mathrm{d} z, \mathrm{d} z^*$ basis.
For partial derivatives, $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial z^*}$ is meant to be the dual basis to $\mathrm{d}z$ and $\mathrm{d}z^*$. That is, we specifically want to have
$$ \frac{\partial z}{\partial z} = 1 \qquad \qquad \frac{\partial z}{\partial z^*} = 0 \\ \frac{\partial z^*}{\partial z} = 0 \qquad \qquad \frac{\partial z^*}{\partial z^*} = 1 $$
Similarly, $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ are the dual basis to $\mathrm{d}x$ and $\mathrm{d}y$, so that we have $$ \frac{\partial x}{\partial x} = 1 \qquad \qquad \frac{\partial x}{\partial y} = 0 \\ \frac{\partial y}{\partial x} = 0 \qquad \qquad \frac{\partial y}{\partial y} = 1 $$
By whatever method you like, we arrive at the equations $$ \frac{\partial x}{\partial z} = \frac{1}{2} \qquad \frac{\partial x}{\partial z^*} = \frac{1}{2} \\ \frac{\partial y}{\partial z} = \frac{1}{2\mathbf{i}} \qquad \frac{\partial y}{\partial z^*} = -\frac{1}{2\mathbf{i}} $$ or alternatively, $$ \frac{\partial z}{\partial x} = 1 \qquad \frac{\partial z}{\partial y} = \mathbf{i} \\ \frac{\partial z^*}{\partial x} = 1 \qquad \frac{\partial z^*}{\partial y} = -\mathbf{i} $$
And we can do linear algebra to express $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial z^*}$ in terms of $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ and vice versa. The end result will be
$$ \frac{\partial}{\partial z} = \frac{1}{2} \frac{\partial}{\partial x} + \frac{1}{2 \mathbf{i}} \frac{\partial}{\partial y} \qquad \qquad \frac{\partial}{\partial z^*} = \frac{1}{2} \frac{\partial}{\partial x} - \frac{1}{2 \mathbf{i}} \frac{\partial}{\partial y} \\ \frac{\partial}{\partial x} = \frac{\partial}{\partial z} + \frac{\partial}{\partial z^*} \qquad \qquad \frac{\partial}{\partial x} = \mathbf{i} \frac{\partial}{\partial z} - \mathbf{i} \frac{\partial}{\partial z^*} $$
we can verify, for example, the formula for $\frac{\partial}{\partial z}$ by applying both sides to, for example, both $z$ and to $z^*$:
$$ \frac{\partial z}{\partial z} = 1 \qquad \qquad \frac{1}{2} \frac{\partial z}{\partial x} + \frac{1}{2 \mathbf{i}} \frac{\partial z}{\partial y} = \frac{1}{2} + \frac{1}{2} = 1$$ $$ \frac{\partial z^*}{\partial z} =0 \qquad \qquad \frac{1}{2} \frac{\partial z^*}{\partial x} + \frac{1}{2 \mathbf{i}} \frac{\partial z^*}{\partial y} = \frac{1}{2} - \frac{1}{2} = 0$$