Expansions of 1/(1 + x + x²) and ln(1 + x + x²)

1. Since $1+x+x^2=\dfrac{1-x^3}{1-x}$,
$$\frac1{1+x+x^2}=\frac{1-x}{1-x^3}=(1-x)\left(1+x^3+x^6+o(x^7)\right)=1-x+x^3-x^4+x^6-x^7+o(x^7).$$
The coefficients repeat with period $3$: $1,-1,0,1,-1,0,\dots$

2. Similarly $\ln(1+x+x^2)=\ln(1-x^3)-\ln(1-x)$, with
$$-\ln(1-x)=x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+o(x^4),\qquad\ln(1-x^3)=-x^3+o(x^4).$$
Hence
$$\ln(1+x+x^2)=x+\frac{x^2}2-\frac{2x^3}3+\frac{x^4}4+o(x^4).$$

Remark. Dividing a fraction of expansions by hand is painful; here a single algebraic identity turns the problem into two geometric series.