[proofplan]
We restrict the [Möbius transformation](/page/M%C3%B6bius%20Transformation) to the open subset of the complex plane on which its finite formula is defined. At an arbitrary point $z$ in this subset, we compute the complex difference quotient directly, which is the quotient-rule calculation written out from first principles. Simplifying the numerator gives the determinant $ad-bc$, and the nondegeneracy condition for a Möbius transformation then implies that the derivative is nonzero.
[/proofplan]
[step:Restrict the finite formula to its complex domain]
Define the finite domain
\begin{align*}
\Omega=\{\zeta \in \mathbb{C}: c\zeta+d \ne 0\}.
\end{align*}
On $\Omega$, the Möbius transformation is represented by the holomorphic quotient
\begin{align*}
T|_{\Omega}: \Omega \to \mathbb{C}, \qquad \zeta \mapsto \frac{a\zeta+b}{c\zeta+d}.
\end{align*}
Let $z \in \Omega$. Since $cz+d \ne 0$, the denominator in the displayed formula is nonzero at $z$.
[/step]
[step:Compute the complex difference quotient at a finite regular point]
Because the map
\begin{align*}
h \mapsto c(z+h)+d
\end{align*}
is continuous at $0$ and has value $cz+d \ne 0$ at $h=0$, there is a radius $\rho>0$ such that $c(z+h)+d \ne 0$ whenever $h \in \mathbb{C}$ and $0<|h|<\rho$. For such $h$, the difference quotient is
\begin{align*}
\frac{T(z+h)-T(z)}{h}
=
\frac{1}{h}
\left(
\frac{a(z+h)+b}{c(z+h)+d}
-
\frac{az+b}{cz+d}
\right).
\end{align*}
Combining the two fractions over the common denominator gives
\begin{align*}
\frac{T(z+h)-T(z)}{h}
=
\frac{(a(z+h)+b)(cz+d)-(az+b)(c(z+h)+d)}
{h(c(z+h)+d)(cz+d)}.
\end{align*}
We simplify the numerator by expanding only the terms depending on $h$:
\begin{align*}
(a(z+h)+b)(cz+d)-(az+b)(c(z+h)+d)
=
(az+b+ah)(cz+d)-(az+b)(cz+d+ch).
\end{align*}
The common term $(az+b)(cz+d)$ cancels, leaving
\begin{align*}
(a(z+h)+b)(cz+d)-(az+b)(c(z+h)+d)
=
h(a(cz+d)-c(az+b)).
\end{align*}
Therefore
\begin{align*}
\frac{T(z+h)-T(z)}{h}
=
\frac{a(cz+d)-c(az+b)}
{(c(z+h)+d)(cz+d)}.
\end{align*}
Since $h \to 0$ implies $c(z+h)+d \to cz+d$, taking the limit gives
\begin{align*}
T'(z)=\frac{a(cz+d)-c(az+b)}{(cz+d)^2}.
\end{align*}
[guided]
We want the derivative at a point $z$ where the finite formula makes sense, so we first make sure the nearby difference quotients are also defined. Since $cz+d \ne 0$ and the function
\begin{align*}
h \mapsto c(z+h)+d
\end{align*}
is continuous at $0$, there exists $\rho>0$ such that $c(z+h)+d \ne 0$ for every $h \in \mathbb{C}$ with $0<|h|<\rho$. Thus the expression $T(z+h)$ is finite for all sufficiently small nonzero $h$.
For those $h$, compute the complex difference quotient from the definition of derivative:
\begin{align*}
\frac{T(z+h)-T(z)}{h}
=
\frac{1}{h}
\left(
\frac{a(z+h)+b}{c(z+h)+d}
-
\frac{az+b}{cz+d}
\right).
\end{align*}
We combine the two fractions using the common denominator $(c(z+h)+d)(cz+d)$, which is nonzero by the choice of $h$ and by $cz+d \ne 0$:
\begin{align*}
\frac{T(z+h)-T(z)}{h}
=
\frac{(a(z+h)+b)(cz+d)-(az+b)(c(z+h)+d)}
{h(c(z+h)+d)(cz+d)}.
\end{align*}
The important cancellation is the same cancellation encoded in the quotient rule. Write $a(z+h)+b=az+b+ah$ and $c(z+h)+d=cz+d+ch$. Then
\begin{align*}
(a(z+h)+b)(cz+d)-(az+b)(c(z+h)+d)
=
(az+b+ah)(cz+d)-(az+b)(cz+d+ch).
\end{align*}
The terms $(az+b)(cz+d)$ cancel. The remaining terms are
\begin{align*}
ah(cz+d)-(az+b)ch
=
h(a(cz+d)-c(az+b)).
\end{align*}
Substituting this into the difference quotient cancels the factor $h \ne 0$:
\begin{align*}
\frac{T(z+h)-T(z)}{h}
=
\frac{a(cz+d)-c(az+b)}
{(c(z+h)+d)(cz+d)}.
\end{align*}
Now the numerator is independent of $h$, and the denominator has an ordinary complex limit because multiplication and addition are continuous in $\mathbb{C}$. Since $c(z+h)+d \to cz+d$ as $h \to 0$, we obtain
\begin{align*}
T'(z)=\lim_{h \to 0}\frac{T(z+h)-T(z)}{h}
=
\frac{a(cz+d)-c(az+b)}{(cz+d)^2}.
\end{align*}
[/guided]
[/step]
[step:Simplify the numerator to the determinant]
The numerator is
\begin{align*}
a(cz+d)-c(az+b)=acz+ad-acz-bc=ad-bc.
\end{align*}
Hence
\begin{align*}
T'(z)=\frac{ad-bc}{(cz+d)^2}.
\end{align*}
[/step]
[step:Use nondegeneracy to prove the derivative never vanishes]
Because $T$ is a Möbius transformation represented by the coefficients $a,b,c,d$, the representing matrix is nondegenerate:
\begin{align*}
ad-bc \ne 0.
\end{align*}
Also $z \in \Omega$, so $cz+d \ne 0$, and therefore
\begin{align*}
(cz+d)^2 \ne 0.
\end{align*}
Thus the quotient
\begin{align*}
\frac{ad-bc}{(cz+d)^2}
\end{align*}
is nonzero. Consequently $T'(z)\ne 0$ for every $z \in \mathbb{C}$ with $cz+d \ne 0$, as claimed.
[/step]
