This is the subgraph. I just wanted to do a couple of exercises with chromatic polynomials, but around the web are pretty standard ones, so i came up with this one myself. I thought it's pretty straightforward.
It is: i color $A$ with $x$ colors, automatically $B$ and $C$ are both different, so $B$ is colored with $x-1$ colors and $C$ with $x-2$ colors and $E$ with $x-1$ colors. This gives us that $F$ must be colored with $x-3$ colors and D with $x-2$ colors.
And it gives me polynomial equal to $$x(x-1)^2(x-2)^2(x-3)$$ But according to wikipedia, coefficient next to $[x^{n-1}]$ is equal to minus number of edges. Wolfram says that $$[x^{n-1}]x(x-1)^2(x-2)^2(x-3) = -9$$ but we have clearly 8 edges here. Seems like i counted something wrong (too much?), so i'd love to hear some hints or solutions to my problem. Thank you in advance!
The problem here is that you haven't accounted for the fact that $E$ and $B$ might have the same color, and $E$ and $C$ might have the same color.
There are cases here that need considering:
Case 1: $E$ has the same color as $B$.
In this case, there are $x$ choices for $A$; $x-1$ for $B$ and $E$; $x-2$ for $C$; $x-1$ for $D$; and $x-2$ for $F$. So, the contribution by this case is $x(x-1)^2(x-2)^2$.
Case 2: $E$ has the same color as $C$.
In this case, there are $x$ choices for $A$; $x-1$ choices for $C$ and $E$; $x-2$ for $B$; $x-2$ for $d$; and $x-1$ for $F$. So, the contribution by this case is $x(x-1)^2(x-2)^2$ as well.
Case 3: $E$, $B$, and $C$ all have different colors.
In this case, there are $x$ choices for $A$, $x-1$ for $B$, $x-2$ for $C$, and $x-3$ for $E$; given these choices, there are $x-2$ choices for $D$ and $x-2$ choices for $F$. So, the contribution by this case is $x(x-1)(x-2)^3(x-3)$.
So, all told, the chromatic polynomial for $G$ is $$ \begin{align*} P_G(x)&=2x(x-1)^2(x-2)^2+x(x-1)(x-2)^3(x-3)\\ &=x(x-1)(x-2)^2(x^2-3x+4)\\ &=x^6-8x^5+27x^4-48x^3+44x^2-16x. \end{align*} $$