[proofplan]
The proof is obtained by applying the ordinary Leibniz rule for the connection and then projecting the resulting $E$-valued one-form to its two type components. The differential of a smooth complex-valued function on a complex manifold decomposes as $df = \partial f + \bar\partial f$. Since the projections to type are $C^\infty(X;\mathbb{C})$-linear on $E$-valued forms, the scalar factor $f$ passes through the projections.
[/proofplan]
[step:Apply the ordinary connection Leibniz rule]
Let $f \in C^\infty(X;\mathbb{C})$ and let $s \in \Gamma(E)$. Since $\nabla$ is a complex connection, it satisfies the Leibniz rule with respect to multiplication by smooth complex-valued functions:
\begin{align*}
\nabla(fs) = df \otimes s + f\nabla s.
\end{align*}
Here $df \otimes s \in A^1(X;E)$ is the $E$-valued one-form obtained by tensoring the complex-valued one-form $df \in A^1(X;\mathbb{C})$ with the section $s \in \Gamma(E)$.
[/step]
[step:Project the Leibniz rule to the $(1,0)$ component]
Let
\begin{align*}
\pi^{1,0}: A^1(X;E) \to A^{1,0}(X;E)
\end{align*}
denote the projection onto the $(1,0)$ component. Applying $\pi^{1,0}$ to the identity from the previous step gives
\begin{align*}
\nabla^{1,0}(fs) = \pi^{1,0}(df \otimes s + f\nabla s).
\end{align*}
The type decomposition of the ordinary differential on a complex manifold gives
\begin{align*}
df = \partial f + \bar\partial f,
\end{align*}
where $\partial f \in A^{1,0}(X)$ and $\bar\partial f \in A^{0,1}(X)$. Therefore
\begin{align*}
\pi^{1,0}(df \otimes s) = \partial f \otimes s.
\end{align*}
Since $\pi^{1,0}$ is $C^\infty(X;\mathbb{C})$-linear on $E$-valued one-forms,
\begin{align*}
\pi^{1,0}(f\nabla s) = f\pi^{1,0}(\nabla s) = f\nabla^{1,0}s.
\end{align*}
Combining these equalities yields
\begin{align*}
\nabla^{1,0}(fs) = \partial f \otimes s + f\nabla^{1,0}s.
\end{align*}
[guided]
We start from the identity already obtained from the connection Leibniz rule:
\begin{align*}
\nabla(fs) = df \otimes s + f\nabla s.
\end{align*}
To isolate the $(1,0)$ part, we use the projection
\begin{align*}
\pi^{1,0}: A^1(X;E) \to A^{1,0}(X;E).
\end{align*}
By definition of the type component of the connection,
\begin{align*}
\nabla^{1,0}(fs) = \pi^{1,0}(\nabla(fs)).
\end{align*}
Substituting the connection Leibniz rule gives
\begin{align*}
\nabla^{1,0}(fs) = \pi^{1,0}(df \otimes s + f\nabla s).
\end{align*}
Now we separate the two summands. The differential of a smooth complex-valued function on a complex manifold decomposes by type as
\begin{align*}
df = \partial f + \bar\partial f,
\end{align*}
with $\partial f$ of type $(1,0)$ and $\bar\partial f$ of type $(0,1)$. Tensoring with $s$ preserves the form type, so $\partial f \otimes s$ is the $(1,0)$ part of $df \otimes s$, while $\bar\partial f \otimes s$ is the $(0,1)$ part. Hence
\begin{align*}
\pi^{1,0}(df \otimes s) = \partial f \otimes s.
\end{align*}
For the second summand, the important point is that projection to type acts only on the form component and is linear over smooth complex-valued functions. Therefore multiplying the $E$-valued one-form $\nabla s$ by the scalar function $f$ commutes with projection:
\begin{align*}
\pi^{1,0}(f\nabla s) = f\pi^{1,0}(\nabla s).
\end{align*}
By the definition $\nabla^{1,0} = \pi^{1,0}\circ\nabla$, this becomes
\begin{align*}
\pi^{1,0}(f\nabla s) = f\nabla^{1,0}s.
\end{align*}
Putting the projected first summand and projected second summand together gives
\begin{align*}
\nabla^{1,0}(fs) = \partial f \otimes s + f\nabla^{1,0}s.
\end{align*}
[/guided]
[/step]
[step:Project the Leibniz rule to the $(0,1)$ component]
Let
\begin{align*}
\pi^{0,1}: A^1(X;E) \to A^{0,1}(X;E)
\end{align*}
denote the projection onto the $(0,1)$ component. Applying $\pi^{0,1}$ to the ordinary Leibniz rule gives
\begin{align*}
\nabla^{0,1}(fs) = \pi^{0,1}(df \otimes s + f\nabla s).
\end{align*}
Using $df = \partial f + \bar\partial f$, the $(0,1)$ component of $df \otimes s$ is $\bar\partial f \otimes s$, so
\begin{align*}
\pi^{0,1}(df \otimes s) = \bar\partial f \otimes s.
\end{align*}
The projection $\pi^{0,1}$ is $C^\infty(X;\mathbb{C})$-linear, hence
\begin{align*}
\pi^{0,1}(f\nabla s) = f\pi^{0,1}(\nabla s) = f\nabla^{0,1}s.
\end{align*}
Combining the two projected terms gives
\begin{align*}
\nabla^{0,1}(fs) = \bar\partial f \otimes s + f\nabla^{0,1}s.
\end{align*}
Together with the $(1,0)$ identity, this proves both asserted Leibniz rules.
[/step]
