Background of the question:
Let $S$ be the family of univalent functions on the unit disk such that $\forall f\in S, f(0)=0, f'(0)=1$.
Loewner's differential equation is about single-slit mappings within $S$ with the expansion of $f_t(z) = e^{-t}(z+\sum_{n=2}^{\infty}a_n(t)z^n)$ with $a_n(t)$ being continuous and $f_0(z) = z$
The actual differential equation is $\frac{\partial f_t}{\partial t} = -f_t\frac{1+k(t)f_t}{1-k(t)f_t}$, where $k(t)$ is a continuous complex-valued function with $|k(t)|=1, 0 \leq t <\infty$. In addiction, $\lim_{t \to \infty}e^tf_t(z)=f(z), \forall |z| < 1$ This convergence is uniform on each compact subset of the unit disk.
Now the distortion theorem states that $\forall f \in S, \frac{1-|z|}{(1+|z|)^3} \leq |f'(z)| \leq \frac{1+|z|}{(1-|z|)^3}$. To show this, the differential equation need to modified by taking derivative w.r.t. z on both side. Let $\frac{\partial f_t}{\partial z} = f'$, and let $f_t = f$ to simplify expression. Then the original equation is equivalent to $\frac{\partial f'}{\partial t} = f'\frac{k^2f^2-2kf-1}{(1-f)^2}$
I figured it would be beneficial to solve the equation first with $k = 1$, which in theory, should produce the derivative of Koebe's function because the original equation produces the Koebe's function when setting $k = 1$. (Koebe's function is $f(z) = \frac{z}{(1-z)^2}$)
My attempt:
Set $k=1$, then $\frac{\partial f'}{\partial t} = f'\frac{f^2-2f-1}{(1-f)^2}$
Rearrange to get $\frac{(1-f)^2}{f'(f^2-2f-1)}\frac{\partial f'}{\partial t} = 1$ Call this equation A.
When $k=1, \frac{\partial f}{\partial f'} = \frac{\frac{\partial f}{\partial t}}{\frac{\partial f'}{\partial t}} = \frac{-f\frac{1+f}{1-f}}{f'\frac{f^2-2f-1}{(1-f)^2}} = \frac{f(f+1)(f-1)}{f'(f^2-2f-1)}$
Note that $(1-f)^2 = 3f(f-1)-f(f+1)-f^2+2f+1$
Then A can be rearranged to ($\frac{3f(f+1)(f-1)}{(f+1)(f^2-2f-1)} - \frac{f(f+1)(f-1)}{(f-1)(f^2-2f-1)} - 1) \frac{1}{f'} \frac{\partial f'}{\partial t} = 1$
$\frac{3}{f+1} \frac{\partial f}{\partial f'}\frac{\partial f'}{\partial t}-\frac{1}{f-1} \frac{\partial f}{\partial f'}\frac{\partial f'}{\partial t}-\frac{1}{f'} \frac{\partial f'}{\partial t} = 1$
$\frac{\partial}{\partial t} (3\ln(f+1))-\frac{\partial}{\partial t}(\ln(f-1))-\frac{\partial}{\partial t}(\ln f') = 1$
$c+3\ln(f+1)-\ln(f-1)-\ln(f') = t$
$e^c = \frac{e^t f'(f-1)}{(f+1)^3}$
substitute $f_0(z) = z$ to get $c = \ln(\frac{z-1}{(z+1)^3})$
Take the limit $t \to \infty$ to obtain $\lim_{t \to \infty}\frac{z-1}{(z+1)^3} \frac{(f_t+1)^3}{f_t-1} = \lim_{t \to \infty} e^tf_t'$
since $\lim_{t \to \infty} f_t = 0,\ -\frac{z-1}{(z+1)^3} = f'(z)$
$f'(z) = \frac{1-z}{(z+1)^3}$
I can't find what has gone wrong, because this is the derivative of a rotated Koebe's function. Please help me identify what has gone wrong.