Partial derivatives chart-dependent? Let $X$ be a Riemann surface. Let $Y\subset X$ be open.
The following definitions are taken from page 60 of Forster's Riemann Surfaces.
We call a function $f\colon Y \rightarrow \mathbb{C}$ (infinitely) differentiable on $Y$ if for every chart $z\colon U \rightarrow V \subset \mathbb{C}$ on $X$ with $U\subset Y$ there exists an (infinitely) real-differentiable function $\tilde{f}\colon V \rightarrow \mathbb{C}$ with $f\restriction_{U}=\tilde{f}\circ z$.
Denote the set of differentiable functions on $Y$ by $\mathcal{E}(Y)$.
Now, let $z_1=x_1+i y_1$ be a chart of $X$ on $U$. With respect to this chart, we define operators $\frac{\partial}{\partial x_1}, \frac{\partial}{\partial y_1}\colon \mathcal{E}(Y)\rightarrow \mathcal{E}(Y)$ by setting $\frac{\partial f}{\partial x_1}\colon = \frac{\partial (f\circ z_1^{-1})}{\partial x}\circ z_1$ and $\frac{\partial f}{\partial y_1}\colon = \frac{\partial (f\circ z_1^{-1})}{\partial y}\circ z_1$ for $f\in \mathcal{E}(Y)$.
Here, for a given $g\colon V\rightarrow \mathbb{C}$, the symbol $\frac{\partial g}{\partial x}$ denotes the usual partial derivative, i.e. $\frac{\partial g}{\partial x}=\frac{\partial}{\partial x} u(x,y) + i \frac{\partial}{\partial x} v(x,y)$ where $g=u(x,y)+i\cdot v(x,y)$.
Now to my question: Is this definition of partial derivatives independent of charts?
 A: No, they depend on the chart.
For instance, let $M = \Bbb C$ be the complex manifold $\Bbb C$ thought abstractly.
Consider the two global holomorphic charts
$$
z_1 \colon z \in M \mapsto z_1(z) = z \in \Bbb C \quad \text{and} \quad
z_2 \colon z \in M \mapsto z_2(z) = 2z \in \Bbb C,
$$
and let's write $z_1 = x_1+ iy_1$ and $z_2 = x_2 + iy_2$.
Let $f\colon M \to \Bbb C$ be the identity map.
Then
$$
\frac{\partial f}{\partial x_1} := \frac{\partial (f\circ z_1^{-1})}{\partial x} = 1,
$$
but
$$
\frac{\partial f}{\partial x_2} := \frac{\partial (f\circ z_2^{-1})}{\partial x} = \frac{1}{2}.
$$
What is independent of the chart is their differentiable properties, and that the partial derivatives satisfy the Cauchy-Riemann equations in one complex chart if and only if they satisfy it in the other.
However, they are related by a simple formula, relying on the chain rule:
$$
\frac{\partial f}{\partial x_2} = \frac{\partial (f\circ z_2^{-1})}{\partial x} 
= \frac{\partial \left((f\circ z_1^{-1}) \circ (z_1\circ z_2^{-1}) \right)}{\partial x} 
= \frac{\partial f}{\partial x_1} \cdot \frac{\partial (z_1\circ z_2^{-1})}{\partial x}.
$$
Indeed, here, $\dfrac{\partial (z_1\circ z_2^{-1})}{\partial x} = \dfrac{\partial (\frac{1}{2}z)}{\partial x} = \dfrac{1}{2}$.
