[proofplan] The composition is well-defined because $f(U)\subset V$ and $g$ is defined on $V$. Since $f$ and $g$ are holomorphic on their domains, the standard composition theorem for holomorphic functions gives that $g\circ f$ is holomorphic on $U$. The complex chain rule gives the derivative formula, and the non-vanishing of $f'$ and $g'$ then implies the derivative of $g\circ f$ is nonzero at every point of $U$. [/proofplan] [step:Define the composed map and verify that it is holomorphic] Define the map \begin{align*} h:U\to W,\qquad z\mapsto g(f(z)). \end{align*} This is well-defined because $f:U\to V$ and $g:V\to W$. Since $f$ is holomorphic on $U$ and $g$ is holomorphic on $V$, and since $f(U)\subset V$, the standard theorem that compositions of holomorphic functions are holomorphic implies that $h=g\circ f$ is holomorphic on $U$. [guided] We first have to check that the expression $g(f(z))$ makes sense for every $z\in U$. The map $f:U\to V$ sends each point $z\in U$ to a point $f(z)\in V$, and the map $g:V\to W$ is defined on every point of $V$. Therefore the composition \begin{align*} h:U\to W,\qquad z\mapsto g(f(z)) \end{align*} is a well-defined function. Next we verify holomorphicity. The function $f$ is holomorphic on $U$ by conformality of $f$, and the function $g$ is holomorphic on $V$ by conformality of $g$. Since the range of $f$ lies in the domain of $g$, the standard composition theorem for holomorphic functions applies to $g\circ f$. It follows that $h=g\circ f$ is holomorphic on $U$. [/guided] [/step] [step:Apply the complex chain rule to compute the derivative] Let $z\in U$. Since $f$ is complex differentiable at $z$ and $g$ is complex differentiable at $f(z)\in V$, the complex chain rule applies to $g\circ f$ at $z$. Hence \begin{align*} h'(z)=(g\circ f)'(z)=g'(f(z))f'(z). \end{align*} [/step] [step:Use non-vanishing derivatives to prove conformality of the composition] Let $z\in U$. Since $f$ is conformal on $U$, we have $f'(z)\neq 0$. Since $f(z)\in V$ and $g$ is conformal on $V$, we also have $g'(f(z))\neq 0$. Therefore the product in the derivative formula is nonzero: \begin{align*} (g\circ f)'(z)=g'(f(z))f'(z)\neq 0. \end{align*} Thus $g\circ f$ is holomorphic on $U$ and has nonzero complex derivative at every point of $U$. By the definition of conformality used in the statement, $g\circ f:U\to W$ is conformal on $U$, with derivative formula as claimed. [/step]