Power serie of $f'/f$ It seems that I'm [censored] blind in searching the power series expansion of $$f(x):=\frac{2x-2}{x^2-2x+4}$$ in $x=0$.
I've tried a lot, e.g., partiell fraction decomposition, or regarding $f(x)=\left(\log((x+1)^2+3)\right)'$ -- without success.  
I' sure that I'm overseeing a tiny little missing link; dear colleagues, please give me a hint.

Given

$$
f(x) = \frac{ 2 x - 2 }{ x^2 - 2 x + 4 }. \tag 1
$$

Assuming you want to expand $f(x)$.
Let
$$
\phi_\pm = 1 \pm \mathtt{i} \sqrt{3}. \tag 2
$$
We can write (1) as
$$
f(x) =
   \frac{ 1 }{ x - \phi_+ } + 
   \frac{ 1 }{ x - \phi_- }. \tag 3
$$
Whence
$$
f(x) =
   - \sum_{k=0}^\infty \left(
      \frac{1}{{\phi_+}^{k+1}} +
      \frac{1}{{\phi_-}^{k+1}}
   \right) x^k. \tag 4
$$
Note that
$$
\phi_\pm = 1 \pm \mathtt{i} \sqrt{3} =
   2 {\exp}\big( \pm \mathtt{i} \pi / 3 \big). \tag 5
$$
Putting (5) in (4) yields

$$
\bbox[16px,border:2px solid #800000] {
   \frac{ 2 x - 2 }{ x^2 - 2 x + 4 } =
      - \sum_{k=0}^\infty {\cos}\big( [k+1] \pi / 3 \big)
      \left( \frac{x}{2} \right)^k.
} \tag {I}
$$

Simple expansion yields

$$
\bbox[16px,border:2px solid #800000] { \frac{ 2 x - 2 }{ x^2 - 2 x + 4 } =
   - \frac{1}{2}
   + \frac{x}{4} + \frac{x^2}{4} + \frac{x^3}{16}
   - \frac{x^4}{32} - \frac{x^5}{32} - \frac{x^6}{128}
   + \cdots
} \tag{II}
$$


Note that
$$
f(x) = \frac{g'(x)}{g(x)}, \tag 6
$$
where
$$
g(x) = x^2 - 2x + 4. \tag 7
$$

If you want to expand
$$
\frac{f'(x)}{f(x)}, \tag 8
$$
then use the simple relation
$$
\frac{f'(x)}{f(x)} = \frac{g''(x)}{g'(x)} - \frac{g'(x)}{g(x)}. \tag 9
$$
Note that
$$
\frac{g''(x)}{g'(x)} = \frac{2x}{2x-2} = - \sum_{k=0}^\infty x^k \tag {10}.
$$
Combination with (II) yields

$$
\bbox[16px,border:2px solid #800000] {
   \frac{f'(x)}{f(x)} =
      + \sum_{k=0}^\infty \left(
         2^{-k} {\cos}\big( [k+1] \pi / 3 \big) - 1
      \right)
      x^k.
} \tag {III}
$$

Simple expansion yields

$$
\bbox[16px,border:2px solid #800000] { \frac{f'(x)}{f(x)} =
   - \frac{1}{2}
   - \frac{5x}{4} - \frac{5x^2}{4} - \frac{17x^3}{16}
   - \frac{31x^4}{32} - \frac{31x^5}{32} - \frac{127x^6}{128}
   \cdots
} \tag{IV}
$$

