<s>[INST] 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
 [/INST] 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.

</s>