= Solution
The three quadratic subfields are $\mathbb Q(\sqrt{-2})$, $\mathbb Q(\sqrt{-3})$ and $\mathbb Q(\sqrt6)$, with respective <fundamental discriminants> $-8,-3,24$. Their associated quadratic <Dirichlet characters> are the <Kronecker symbols>
$$
\chi_{-8}(n)=\left(\frac{-8}{n}\right),\qquad
\chi_{-3}(n)=\left(\frac{-3}{n}\right),\qquad
\chi_{24}(n)=\left(\frac{24}{n}\right).
$$
The <regular representation> of $C_2\times C_2$ is the direct sum of the trivial character and these three nontrivial characters. At an unramified prime the character is the sign by which <Frobenius automorphism> acts on the square root defining its quadratic subfield; at a ramified prime the <linear character> has no <inertia group> invariants. The <local factor of an Artin L-function> consequently gives
$$
L(\rho,s)=\zeta(s)L(\chi_{-8},s)L(\chi_{-3},s)L(\chi_{24},s).
$$
The zero values of a quadratic <Dirichlet character> at conductor primes mean that its local factor is one, not that the whole <Artin L-function> vanishes there.
Only the rational primes $2,3,5,7$ can contribute to coefficients with index at most ten. Their character values are
$$
\begin{array}{c|rrr|l}
p&\chi_{-8}(p)&\chi_{-3}(p)&\chi_{24}(p)&L_p(\rho,s)\ \text{with }T=p^{-s}\\ \hline
2&0&-1&0&(1-T^2)^{-1}\\
3&1&0&0&(1-T)^{-2}\\
5&-1&-1&1&(1-T^2)^{-2}\\
7&-1&1&-1&(1-T^2)^{-2}
\end{array}
$$
For the odd unramified primes these entries follow by checking squares: modulo five, both $-8$ and $-3$ are the nonsquare $2$, while $24$ is the square $4$; modulo seven they are respectively $6,4,3$. At $3$, $-8\equiv1$ is a square. At $2$, the $-3$ character is $-1$ because the integral quadratic generator has polynomial $X^2+X+1$, irreducible modulo two. The other entries at $2$ and $3$ are zero exactly where the corresponding <fundamental discriminant> is divisible by that prime. This verifies every local factor, including the ramified ones.
The <Euler product> makes $a_n$ a <multiplicative arithmetic function>. Expanding the first two factors gives
$$
(1-T^2)^{-1}=1+T^2+T^4+\cdots,\qquad
(1-T)^{-2}=1+2T+3T^2+\cdots.
$$
Therefore $a_2=0$, $a_4=1$, $a_8=0$, $a_3=2$ and $a_9=3$. Both factors for $5$ and $7$ have zero coefficient of $T$, so $a_5=a_7=0$. Multiplicativity gives $a_6=a_2a_3=0$ and $a_{10}=a_2a_5=0$, while $a_1=1$. The answer is
$$
\boxed{(a_1,\ldots,a_{10})=(1,0,2,1,0,0,0,0,3,0)}.
$$
As an arithmetic check, the <regular representation> <Artin L-function> is the <Dedekind zeta function> of $F$. The local factors say there is one prime of norm four above two and two primes of norm three above three. Their powers give respectively one ideal of norm four and three ideals of norm nine, matching the nonzero coefficients above.
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