find taylor series to fourth term I'm wondering if there is faster method than just calculating derivatives with finding taylor series up to 4 term of function $\displaystyle f(x)=\frac{(1+x^4)}{(1+2x)^3(1-2x)^2}$  

Since the Taylor series of $g(x)=\frac{x^4}{(1+2x)^3(1-2x)^2}$ in a neighbourhood of the origin is given by $x^4+o(x^4)$, it is enough to compute the Taylor series of:
$$\begin{eqnarray*} h(x)&=&\frac{1}{(1+2x)^3(1-2x)^2}\\&=&\left(1-6 x+24 x^2-80 x^3+240 x^4+o(x^4)\right)\left(1+4 x+12 x^2+32 x^3+80 x^4+o(x^4)\right)\\&=&1 - 2 x + 12 x^2 - 24 x^3 + 96 x^4+o(x^4)\tag{1}\end{eqnarray*}$$
giving:
$$ \frac{1+x^4}{(1+2x)^3(1-2x)^2}=1 - 2 x + 12 x^2 - 24 x^3 + 9\color{red}{7} x^4+o(x^4).\tag{2}$$
In $(1)$ we just used the Taylor series for $\frac{1}{(1-z)^m}$, given by:
$$\frac{1}{(1-z)^m}=\sum_{n\geq 0}\binom{n+m-1}{n}\, z^n\tag{3}$$
and took the Cauchy product of two such series. A different approach is given by:
$$ \frac{1}{(1-4x^2)^2}=1+8 x^2+48 x^4+o(x^4),\tag{4} $$
that just leads to $(1)$ since:
$$ \frac{1}{1+2x}=1 - 2 x + 4 x^2 - 8 x^3 + 16 x^4+o(x^4).\tag{5} $$
