conformal map of rectangle to unit disk has first derivative zero

Let $f$ be a conformal map from rectangle $R$ to unit disk. Prove that $f$ analytic at all vertices of $R$ (say $z_k, k=1,2,3,4$), and near $z_k$, $$f(z)=f(z_k)+(z-z_k)^2 h(z)$$ where $h(z)$ is analytic at $z_k$ and non zero.

My attempt: I tried to use Schwarz-Christoffel formula to get a map from rectangle to upper half plane and use Möbius to unit disk, but the first map can't guarantee the analytic of vertices. Thanks for any help.

You don't need the explicit form of $f$. The Schwarz reflection principle is enough. Let's say $R=(0,a)\times (0,b)$ and we study the behavior at $0$. Consider the reflected rectangles $R_1=(-a,0)\times (0,b)$, $R_2=(-a,0)\times (-b,0)$ and $R_3 =(0,a)\times (-b,0)$.
Let $\gamma_1=\partial R \cap \partial R_1$, the vertical line segment from $0$ to $ib$. Its image under $f$ is an arc of the unit circle, call it $\Gamma_1$. By reflection across $\gamma_1$, $f$ extends to a holomorphic map from $R\cup R_1\cup \gamma_1$ to the domain obtained by taking the union of the unit disk, its exterior, and the arc $\Gamma_1$.
Reflecting two more times, we extend $f$ to a holomorphic function $F$ in $(-a,a)\times (-b,b)$. This shows the analyticity at $0$. Also, keeping track of what the reflection was, you'll see that $F(-z)=F(z)$. Indeed, going from $z$ to $-z$ means reflecting in two sides of a rectangle, which in the image plane involves going $w \mapsto 1/\bar w \mapsto 1/\overline{(1/\bar w)} = w$.
Since $F$ is even, its expansion at $0$ has the form $F(z) = \sum_{k=0}^\infty c_{2k} z^{2k}$ which meets the stated requirement (and more).