<s>[INST] Schwarzian derivative as the derivative of the logarithm. There is one interesting fact in Wikipedia article on Schwarzian derivative, namely, if $F(z, w)=\log{\frac{f(z)-f(w)}{z-w}}$ then (sign error fixed) $6\frac{\partial^2 F(z, w)}{\partial z \partial w}|_{z=w} = (Sf)(w).$ I cannot see it. We have $\frac{\partial^2 F(z, w)}{\partial z \partial w} = \frac{f'(w)f'(z)}{(f(z)-f(w))^2}-\frac{1}{(z-w)^2}.$ When $|z-w|$ is small both of summands are large so none of them has no limit but their difference has. I brought them to a common denominator, but nothing happened. How to compute this limit?
 [/INST] This is “just” an exercise in developing Taylor series.
Consider a fixed point $w$ where $f(w) \ne \infty$ and $f'(w) \ne 0$.
Then
$$
 f(z) = a_0 + a_1(z-w) + a_2(z-w)^2 + a_3 (z-w)^3  + \ldots
$$
where $a_n = \frac{f^{(n)}(w)}{n!}$, and $\ldots$ indicates “higher-order
terms of $(z-w)$” to shorten the notation.
Now
$$
 \frac{1}{f(z)-f(w)} = \frac{1}{a_1(z-w) + a_2(z-w)^2 + a_3 (z-w)^3  + \ldots} \\
 = \frac{1}{a_1(z-w)(1 + \frac{a_2}{a_1}(z-w)+ \frac{a_3}{a_1}(z-w)^2 + \ldots)} \\
= \frac{1}{a_1(z-w)} \left( 1 - \frac{a_2}{a_1}(z-w) + \bigl( \frac{a_2^2}{a_1^2} - \frac{a_3}{a_1} \bigr)(z-w)^2 + \ldots \right)
$$
and therefore
$$
\frac{1}{(f(z)-f(w))^2} =  \frac{1}{a_1^2(z-w)^2} \left( 1 - 2\frac{a_2}{a_1}(z-w) + \bigl( 3\frac{a_2^2}{a_1^2} - 2\frac{a_3}{a_1} \bigr)(z-w)^2 + \ldots \right)
$$
On the other hand,
$$
 f'(w)f'(z) = a_1 (a_1 + 2a_2(z-w) + 3a_3(z-w)^2 + \ldots) \\
 = a_1^2 (1 + 2\frac{a_2}{a_1}(z-w) + 3\frac{a_3}{a_1}(z-w)^2 + \ldots) 
$$
so that
$$
 \frac{f'(w)f'(z)}{(f(z)-f(w))^2} = \frac{1}{(z-w)^2} \left( 1 +  \bigl( -\frac{a_2^2}{a_1^2} +\frac{a_3}{a_1} \bigr)(z-w)^2 + \ldots \right)
 = \frac{1}{(z-w)^2} + \bigl( -\frac{a_2^2}{a_1^2} +\frac{a_3}{a_1} \bigr) + \ldots
$$
which gives the desired
$$
\lim_{z \to w} \left( \frac{f'(w)f'(z)}{(f(z)-f(w))^2} - \frac{1}{(z-w)^2} \right)
 = \frac{a_3}{a_1} - \frac{a_2^2}{a_1^2}
 = \frac 16 \frac{f'''(w)}{f'(w) } - \frac 14 \left(\frac{f''(w)}{f'(w)}\right)^2 = \frac 16 (Sf)(w)
$$
Finally, due to the identity theorem, this relation holds everywhere
and not only at points where $f'(w) \ne 0$. 
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