In the following, we consider polynomials in $\mathbb{F}_2[x]$. This means all coefficients are taken modulo $2$. Let $f(n)$ be the number of odd coefficients in $(x^2 - x + 1)^n$.

Claim 1. We have $(x^2 - x + 1)^{2k} = x^{2k+1} - x^{2k} + 1$ for any nonnegative integer $k$.
Proof. It suffices to note that
$$(x^2 - x + 1)^2 = x^4 - 2x^3 + 3x^2 - 2x + 1 = x^4 - x^2 + 1.$$
The result follows easily by induction. $\square$

Claim 2. We have $f(2ka+b) = f(a)f(b)$ for any positive integers $a, b$ satisfying $b < 2^{k-1}$.
Proof. By claim 1, we have
$$(x^2 - x + 1)^{2ka+b} = ((x^2 - x + 1)^{2k})^a (x^2 - x + 1)^b = (x^{2k+1} - x^{2k} + 1)^a (x^2 - x + 1)^b.$$
$$(x^{2k+1} - x^{2k} + 1)^a = \sum_{i=1}^{s} c_i x^{2k \alpha_i},$$
$$(x^2 - x + 1)^b = \sum_{j=1}^{t} d_j x^{\beta_j}.$$
Then their product is
$$\sum_{i=1}^{s} \sum_{j=1}^{t} c_i d_j x^{2k\alpha_i + \beta_j}.$$
Since we have $\beta_j \le 2b < 2^k$, different pairs $(i, j)$ correspond to different exponents $2^k\alpha_i + \beta_j$. Also, $c_i d_j$ is odd if and only if both $c_i$ and $d_j$ are odd. As there are $f(a)$ odd coefficients $c_i$ and $f(b)$ odd coefficients $d_j$, the number of odd coefficients is $f(a)f(b)$. $\square$

Claim 3. We have $f(2^k - 1) = \frac{2^{k+2} - (-1)^k}{3}$ for any positive integer $k$.
Proof. It suffices to make the following observations. For odd $k$, the coefficients of $(x^2 - x + 1)^{2^k-1}$ follow the pattern
$$\underbrace{110110\cdots110}_{\frac{2^{k-2}}{3} \text{ triples } 110} 111 \underbrace{011011\cdots011}_{\frac{2^{k-2}}{3} \text{ triples } 011}$$
while for even $k$, the coefficients of $(x^2 - x + 1)^{2^k-1}$ follow the pattern
$$\underbrace{110110\cdots110}_{\frac{2^{k-1}}{3} \text{ triples } 110} 1 \underbrace{011011\cdots011}_{\frac{2^{k-1}}{3} \text{ triples } 011}.$$
It is not hard to prove these by induction and claim 1, since $2^{k+1}-1 = (2^k-1)2^k$. We omit the details. $\square$

Now, by the above claims, we obtain
$$\begin{aligned} f(2009) &= f(2^6 \times 31 + 25) = f(31)f(25) \\ &= f(31)f(2^3 \times 3 + 1) = f(31)f(3)f(1) \\ &= \frac{2^7 + 1}{3} \cdot \frac{2^4 - 1}{3} \cdot \frac{2^3 + 1}{3} = 645. \end{aligned}$$