For the first question, it may be useful to simplify a little, and observe that $\frac{1+x+x^2}{1-x+x^2}=\frac{2x}{1-x+x^2}+1.$ Note that $\frac{2x}{1-x+x^2}=\frac{2x(1+x)}{1+x^3}.$ The series expansion of $\frac{1}{1+x^3}$ is standard: use the fact that $\frac{1}{1-t}=1+t+t^2+\cdots,$ and set $t=-x^3$. Unfortunately, this will give us a general formula for the $n$-th derivative. But we can refuse to notice that, and just calculate for $n=4$.
We have $\frac{1}{1+x^3}=1-x^3+x^6-x^9+\cdots$. So the $x^4$ term in the power series expansion of our original function is just $-2x^4$.