Let $M$ be a complex manifold and let $f \in C^{\infty}(M)$. Let $\{\partial_i\}$ denote the standard local frame on the tangent bundle. Let $(\phi,U)$ be a chart on $M$ with $p \in U$. Is the following true in general?
\begin{equation}
\partial_{z_i}(f \circ \phi^{-1})(\phi(p))=(d_p f)(\partial_i(p))
\end{equation}
If this is not the case, is there some other relationship between partial derivatives and the differential?
Best Answer
Let me denote by $\varphi_*\colon T_pM\to T_{\varphi(p)}N$ the pushforward of $\varphi\colon M\to N$ and for a function $f\colon M\to \mathbb C$, let $df\colon M\to T^*M$ be its differential.
If $(U,\phi)$, $\phi = (z_1,\ldots,z_n)$ is a chart, we have that $\partial_{z_i}|_{\phi(p)} = \phi_*(\partial_i(p))$. Then,
\begin{align}\partial_{z_i}(f\circ\phi^{-1})(\phi(p)) &= \phi_*(\partial_i(p))(f\circ \phi^{-1})\\ &= (\partial_i(p))(f\circ\phi^{-1}\circ\phi)\\ &= (\partial_i(p))(f)\\ &= (df_p)(\partial_i(p)) \end{align}
where the last equality is true since by definition $(df_p)(X) = X(f)$, for any $X\in T_pM$.
Perhaps instead of considering covector field (differential $1$-form) $df$, you wanted to see the connection with pushforward $f_*\colon T_pM\to T_{f(p)}\mathbb C.$ For that we need to elaborate on the isomorphism $T_q\mathbb C\cong\mathbb C$.
In general, for vector space $V$, we have isomorphism $V\to T_qV$ that sends vector $v$ to directional derivative defined by $v$: $g\mapsto \frac d{dt}g(p + tv)|_{t=0}.$ In the case when $V = \mathbb C$, we can easily write it's inverse as $X\mapsto X(\operatorname{id_{\mathbb C}})\colon T_q\mathbb C\to\mathbb C.$ All in all, for any $X\in T_pM$ we have
$$f_*(X)(\operatorname{id}_{\mathbb C}) = X(\operatorname{id}_{\mathbb C}\circ f) = Xf = (df_p)(X).$$