Let $G\subseteq \mathbb{C}$ be open and let $$f(x,z):X\times G\to \mathbb{C}$$ is holomorphic for each fixed $x\in X$. I want to show that $$\frac{1}{h}(f(x,z_0+h)-f(x,z_0))=\int_0^1\frac{\partial f}{\partial z}(x,z_0+ht)dt, \quad 0<|h|<r,\, x\in X$$
$$h\int_0^1\frac{\partial f}{\partial z}(x,z_0+ht)dt=\int_0^1f'(x,z_0+ht)hdt=\int_\gamma f'(x,w)dw=f(x,z_0+h)-f(x,z_0)$$ where $\gamma=z_0+ht \quad t\in [0,1]$
Notice that $r$ is important because to be holomorphic is a local property, so we want to be inside $G$ when we're working with continuity and derivatives of $f$. Fix $(x,z_0)\in X\times G$, with $G\subset\mathbb{C}$ open. Then we may find $r$ such that $z_0+h\in G$ with $0<|h|<r$. When you write $\frac{\partial}{\partial z}f(x,z_0+ht)$, I think you really mean $\frac{\partial}{\partial z}f(x,z(t))$, where $z(t)=z_0+ht$. In this case we have \begin{align*} \int_0^1\frac{d}{dt}f(x,z(t))dt=f(x,z(1))-f(x,z(0))=f(x,z_0+h)-f(x,z_0). \end{align*} However, $\frac{d}{dt}f(x,z(t))=\frac{\partial}{\partial z}f(x,z(t))\frac{dz}{dt}=h\frac{\partial}{\partial z}f(x,z(t))$. Thus, \begin{align*} f(x,z_0+h)-f(x,z_0)=\int_0^1\frac{d}{dt}f(x,z(t))dt=h\int_0^1\frac{\partial}{\partial z}f(x,z(t))dt. \end{align*}