Have I correctly taken this function $f(x) = \frac{x^{2}}{1+2x}$ and $f ' (x) $and turned them into power series? $f(x) = \dfrac{x^{2}}{1+2x}$
To turn this into a power series I recall the similar looking geometric series,
$f(x) = \sum\limits_{n=1}^{\infty} x^{n} = \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^5 +\cdots$
And not I can mutate my original function to fit the form the geometric series:
$f(x) = \dfrac{x^{2}}{1-(-2x)}$
$f(x) = \dfrac{x^2}{1}*\dfrac{x^{2}}{1-(-2x)} = 1 + (-2x) + (-2x)^{2} + (-2x)^{3} + (-2x)^{4} +\cdots$
$f(x) = \dfrac{x^2}{1}*\dfrac{x^{2}}{1-(-2x)} = 1 - 2x + 4x^2 - 8x^3 + 16x^4 +\cdots$
$f(x) = \dfrac{x^2}{1}*\dfrac{x^{2}}{1-(-x^{2})} = x^2( 1 - 2x + 4x^2 - 8x^3 + 16x^4 +\cdots)$
$f(x) = \dfrac{x^2}{1}*\dfrac{x^{2}}{1-(-x^{2})} = x^2 - 2x^3 + 4x^4 -8x^5 + 16x^{6} +\cdots)$
And for $'(x)$
$f'(x) = 2x - 6x^2 + 16x^3 -40x^4 + 16*6x^{5} +\cdots$
Is this right or wrong?
Thank you
 A: Suggestion:
$$
\frac{1}{2x+1} \approx 1 - 2 x + 4 x^2 - 8 x^3 + 16 x^4 - 32 x^5 +\mathcal{O}\left( x^{6} \right)
$$
Now multiply by $x^{2}$.
$$
f(x) = \frac{x^2}{2 x+1} \approx x^2 -2 x^3+4 x^4-8 x^5+ \mathcal{O}\left( x^{6} \right)
$$

Suggestion:
$$ 
\frac{1}{(2 x+1)^2} \approx 1 - 4 x + 12 x^2 - 32 x^3 + 80 x^4 - 192 x^5 + \mathcal{O}\left( x^{6} \right)
$$
Multiply by $2 x (1 + x) = 2 x^2+2 x$:
$$
f'(x) = \frac{2 x (x+1)}{(2 x+1)^2} \approx 2 x - 6 x^2 + 16 x^3 - 40 x^4 + 96 x^5+ \mathcal{O}\left( x^{6} \right)
$$


@law-of-fives suggests partial fraction decomposition:

$$
f(x) =\frac{x^2}{2 x+1} = \frac{x}{2}+\frac{1}{4 (2 x+1)}-\frac{1}{4}
$$
A more elegant solution!
A: Your approach based upon the geometric series expansion 
\begin{align*}
\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots
\end{align*}
is fine  and appropriate. Just the calculations are not correct in all aspects.

We obtain
  \begin{align*}
f(x)&=\frac{x^2}{1+2x}\\
&=x^2\cdot\frac{1}{1-(-2x)}\\
&=x^2\left(1-2x+4x^2-8x^3+16x^4-\cdots\right)\\
&=x^2-2x^3+4x^4-8x^4+16x^6-\cdots\tag{1}
\end{align*}
and  from (1)  we derive  the  derivative of $f$
\begin{align*}
f^\prime(x)=2x-6x^2+16x^3-32x^3+72x^5-\cdots
\end{align*}
in accordance with your calculation.

