[proofplan]
We first reduce to a homogeneous form of type $(p,q)$, because every smooth complex-valued form has a finite type decomposition. The key points are that a holomorphic map pulls $(p,q)$-forms back to $(p,q)$-forms, and that smooth pullback commutes with the [exterior derivative](/theorems/1525). Once these are established in local holomorphic coordinates, we compare the $(p+1,q)$ and $(p,q+1)$ components of the identity $F^*(d\beta)=d(F^*\beta)$ for a homogeneous form $\beta \in \Omega^{p,q}(Y)$.
[/proofplan]
[step:Check in holomorphic coordinates that pullback preserves bidegree]
Let $m = \dim_{\mathbb{C}} X$ and $n = \dim_{\mathbb{C}} Y$. Fix a point $x_0 \in X$. Choose a holomorphic coordinate chart $(U,z)$ on $X$ with $x_0 \in U$, where
\begin{align*}
z: U \to z(U) \subset \mathbb{C}^m
\end{align*}
has coordinate functions $z_1,\dots,z_m$, and choose a holomorphic coordinate chart $(V,w)$ on $Y$ with $F(x_0) \in V$ and, after shrinking $U$ if necessary, $F(U) \subset V$. Write the component functions of $F$ in these coordinates as smooth complex-valued maps
\begin{align*}
F_a := w_a \circ F|_U: U \to \mathbb{C}
\end{align*}
for $a \in \{1,\dots,n\}$.
Since $F$ is holomorphic, each $F_a$ is holomorphic in the variables $z_1,\dots,z_m$, hence
\begin{align*}
\frac{\partial F_a}{\partial \bar z_i} = 0
\end{align*}
for every $i \in \{1,\dots,m\}$ and $a \in \{1,\dots,n\}$. Therefore
\begin{align*}
F^*(dw_a) = d(F_a) = \sum_{i=1}^{m} \frac{\partial F_a}{\partial z_i}\, dz_i.
\end{align*}
Similarly, since $\overline{F_a}$ is anti-holomorphic,
\begin{align*}
F^*(d\bar w_a) = d(\overline{F_a}) = \sum_{i=1}^{m} \frac{\partial \overline{F_a}}{\partial \bar z_i}\, d\bar z_i.
\end{align*}
Thus $F^*(dw_a)$ has type $(1,0)$ and $F^*(d\bar w_a)$ has type $(0,1)$.
Now let $\beta \in \Omega^{p,q}(Y)$ be a smooth form of pure type $(p,q)$. On $V$ it is a finite sum of coordinate monomials
\begin{align*}
h_{I,J}\, dw_{i_1} \wedge \cdots \wedge dw_{i_p} \wedge d\bar w_{j_1} \wedge \cdots \wedge d\bar w_{j_q},
\end{align*}
where each coefficient is a smooth function
\begin{align*}
h_{I,J}: V \to \mathbb{C},
\end{align*}
and $I=(i_1,\dots,i_p)$, $J=(j_1,\dots,j_q)$ are increasing multi-indices. Pulling back such a monomial gives
\begin{align*}
(h_{I,J}\circ F)\, F^*(dw_{i_1}) \wedge \cdots \wedge F^*(dw_{i_p}) \wedge F^*(d\bar w_{j_1}) \wedge \cdots \wedge F^*(d\bar w_{j_q}).
\end{align*}
The first $p$ factors have type $(1,0)$ and the last $q$ factors have type $(0,1)$, so each summand has type $(p,q)$. Hence
\begin{align*}
F^*\beta \in \Omega^{p,q}(X).
\end{align*}
[guided]
We need the holomorphicity of $F$ precisely to make sure pullback does not mix the two types of one-forms. Fix $x_0 \in X$, choose holomorphic coordinates $z_1,\dots,z_m$ near $x_0$ on $X$, and choose holomorphic coordinates $w_1,\dots,w_n$ near $F(x_0)$ on $Y$. After shrinking the coordinate neighbourhood in $X$, the image lies in the coordinate neighbourhood in $Y$. Define the coordinate component maps
\begin{align*}
F_a := w_a \circ F|_U: U \to \mathbb{C}
\end{align*}
for $a \in \{1,\dots,n\}$.
Because $F$ is holomorphic, each component $F_a$ is holomorphic as a function of the variables $z_1,\dots,z_m$. Therefore its anti-holomorphic partial derivatives vanish:
\begin{align*}
\frac{\partial F_a}{\partial \bar z_i} = 0
\end{align*}
for all $i \in \{1,\dots,m\}$ and $a \in \{1,\dots,n\}$. Computing the pullback of a basic $(1,0)$-form gives
\begin{align*}
F^*(dw_a) = d(w_a \circ F) = d(F_a).
\end{align*}
Expanding the exterior derivative in the chosen complex coordinates and using $\partial F_a/\partial \bar z_i=0$, we get
\begin{align*}
F^*(dw_a) = \sum_{i=1}^{m} \frac{\partial F_a}{\partial z_i}\, dz_i.
\end{align*}
This is a linear combination only of the $(1,0)$ one-forms $dz_i$, so it has type $(1,0)$.
The conjugate coordinate form behaves in the complementary way. Since $\overline{F_a}$ is anti-holomorphic,
\begin{align*}
F^*(d\bar w_a) = d(\overline{w_a \circ F}) = d(\overline{F_a}) = \sum_{i=1}^{m} \frac{\partial \overline{F_a}}{\partial \bar z_i}\, d\bar z_i.
\end{align*}
This is a linear combination only of the $(0,1)$ one-forms $d\bar z_i$, so it has type $(0,1)$.
Now take a pure type form $\beta \in \Omega^{p,q}(Y)$. In the coordinate chart on $Y$, it is a finite sum of terms of the form
\begin{align*}
h_{I,J}\, dw_{i_1} \wedge \cdots \wedge dw_{i_p} \wedge d\bar w_{j_1} \wedge \cdots \wedge d\bar w_{j_q},
\end{align*}
where
\begin{align*}
h_{I,J}: V \to \mathbb{C}
\end{align*}
is smooth. Pullback sends the coefficient to $h_{I,J}\circ F$, sends each $dw_{i_r}$ to a $(1,0)$-form, and sends each $d\bar w_{j_s}$ to a $(0,1)$-form. Therefore the wedge product still has exactly $p$ factors of type $(1,0)$ and exactly $q$ factors of type $(0,1)$. Thus every local summand of $F^*\beta$ has type $(p,q)$, and hence
\begin{align*}
F^*\beta \in \Omega^{p,q}(X).
\end{align*}
[/guided]
[/step]
[step:Use naturality of the exterior derivative on a pure type form]
Let $\beta \in \Omega^{p,q}(Y)$ be a smooth form of pure type $(p,q)$. Since every holomorphic map of complex manifolds is smooth, $F$ has an ordinary smooth pullback on differential forms. By naturality of the exterior derivative under smooth pullback, applied to the smooth map $F: X \to Y$ and the smooth form $\beta$, we have
\begin{align*}
F^*(d\beta) = d(F^*\beta).
\end{align*}
Since $d = \partial + \bar{\partial}$ on complex-valued forms and $\beta$ has type $(p,q)$,
\begin{align*}
d\beta = \partial\beta + \bar{\partial}\beta,
\end{align*}
with
\begin{align*}
\partial\beta \in \Omega^{p+1,q}(Y)
\end{align*}
and
\begin{align*}
\bar{\partial}\beta \in \Omega^{p,q+1}(Y).
\end{align*}
By the bidegree preservation proved above,
\begin{align*}
F^*(\partial\beta) \in \Omega^{p+1,q}(X)
\end{align*}
and
\begin{align*}
F^*(\bar{\partial}\beta) \in \Omega^{p,q+1}(X).
\end{align*}
Also $F^*\beta \in \Omega^{p,q}(X)$, so
\begin{align*}
d(F^*\beta) = \partial(F^*\beta) + \bar{\partial}(F^*\beta),
\end{align*}
with
\begin{align*}
\partial(F^*\beta) \in \Omega^{p+1,q}(X)
\end{align*}
and
\begin{align*}
\bar{\partial}(F^*\beta) \in \Omega^{p,q+1}(X).
\end{align*}
Substituting these decompositions into $F^*(d\beta)=d(F^*\beta)$ gives
\begin{align*}
F^*(\partial\beta) + F^*(\bar{\partial}\beta) = \partial(F^*\beta) + \bar{\partial}(F^*\beta).
\end{align*}
The decomposition of complex-valued forms into bidegree components is direct. Comparing the $(p+1,q)$ components gives
\begin{align*}
F^*(\partial\beta) = \partial(F^*\beta),
\end{align*}
and comparing the $(p,q+1)$ components gives
\begin{align*}
F^*(\bar{\partial}\beta) = \bar{\partial}(F^*\beta).
\end{align*}
[/step]
[step:Sum the identities over the type decomposition of an arbitrary form]
Let $\alpha \in \Omega^\bullet(Y;\mathbb{C})$ be arbitrary. Since $Y$ is a complex manifold, $\alpha$ has a finite type decomposition
\begin{align*}
\alpha = \sum_{p,q} \alpha^{p,q},
\end{align*}
where each component satisfies
\begin{align*}
\alpha^{p,q} \in \Omega^{p,q}(Y).
\end{align*}
Using the pure type identity from the previous step and linearity of $F^*$, $\partial$, and $\bar{\partial}$, we obtain
\begin{align*}
F^*(\partial\alpha) = F^*\left(\sum_{p,q}\partial\alpha^{p,q}\right) = \sum_{p,q}F^*(\partial\alpha^{p,q}) = \sum_{p,q}\partial(F^*\alpha^{p,q}) = \partial\left(\sum_{p,q}F^*\alpha^{p,q}\right) = \partial(F^*\alpha).
\end{align*}
For the $\bar{\partial}$ identity, the finite type decomposition gives
\begin{align*}
\bar{\partial}\alpha = \sum_{p,q}\bar{\partial}\alpha^{p,q}.
\end{align*}
Applying linearity of $F^*$ and the pure type identity $F^*(\bar{\partial}\alpha^{p,q}) = \bar{\partial}(F^*\alpha^{p,q})$ to each summand gives
\begin{align*}
F^*(\bar{\partial}\alpha) = \sum_{p,q}F^*(\bar{\partial}\alpha^{p,q}).
\end{align*}
Therefore
\begin{align*}
F^*(\bar{\partial}\alpha) = \sum_{p,q}\bar{\partial}(F^*\alpha^{p,q}).
\end{align*}
Since the sum is finite and $\bar{\partial}$ is linear,
\begin{align*}
\sum_{p,q}\bar{\partial}(F^*\alpha^{p,q}) = \bar{\partial}\left(\sum_{p,q}F^*\alpha^{p,q}\right).
\end{align*}
Finally, linearity of pullback gives $\sum_{p,q}F^*\alpha^{p,q}=F^*\alpha$, hence
\begin{align*}
F^*(\bar{\partial}\alpha) = \bar{\partial}(F^*\alpha).
\end{align*}
Thus pullback by a holomorphic map commutes with both Dolbeault operators on every smooth complex-valued differential form.
[/step]
