Let's write
$$f(x)=x^6-2x^4+2x^2-2.$$
We immediately notice (see also lhf's answer) that $f(x)=g(x^2)$ where $g(x)=x^3-2x^2+2x-2.$ Furthermore, by Eisenstein's criterion ($p=2$) both $f(x)$ and $g(x)$ are irreducible in $\Bbb{Q}[x]$.
Because the quadratic $g'(x)=3x^2-4x+2$ has a negative discriminant, we can conclude that $g(x)$ is an everywhere increasing function of a real variable $x$. Consequently it has a single real zero and two complex ones. This means that adjoining the real zero of $g(x)$ won't give its splitting field. Therefore the splitting field $K$ of $g(x)$ must be a degree six extension of $\Bbb{Q}$.
If the roots of $g(x)$ in $\Bbb{C}$ are $y_1,y_2,y_3$, then the roots of $f(x)$ are $x_i=\sqrt{y_i}$, $i=1,2,3,$ (for some branch of the complex square root) and $x_{i+3}=-x_i, i=1,2,3$. If $L=\Bbb{Q}(x_1,\ldots,x_6)$ is the splitting field of $f(x)$ inside $\Bbb{C}$, then we can identify the Galois group $G=Gal(L/\Bbb{Q})$ with a subgroup of permutations of the roots, so $G\le S_6$. Obviously not all permutatios will occur as any automorphism $\tau\in G$ must observe the relations $x_{i+3}=-x_i$. Because $K$ is a normal extension of $\Bbb{Q}$, we know that $H=Gal(L/K)\unlhd G$. Furthermore, $G/H\simeq Gal(K/\Bbb{Q})\simeq S_3$.
To make further progress I invoke Dedekind's theorem. For proofs see this on our site and a proof by Tate. I also need help from Mathematica to factor $f(x)$ modulo a few primes.
Let's first figure out $[L:K]$. Clearly
$$L=K(\sqrt{y_1},\sqrt{y_2},\sqrt{y_3}),$$ and adjoining those square roots in sequence either doubles the extension degree or doesn't do anything. Hence $[L:K]\in\{1,2,4,8\}$.
- Modulo $p=3$ the polynomial $f(x)$ actually remains irreducible. This means that there exists a 6-cycle $\sigma\in G$. As $H$ is a 2-group, and $G/H\simeq S_3$, we see that the restriction of $\sigma$ to $K$ must have order three. Hence $\sigma^3\in H$, which easily implies that $\sigma^3(x_i)=-x_i$ for all $i$.
- It can be seen that modulo $p=13$ we have the factorization
$$f(x)=(x+4)(x+9)(x^5+x^2+5).$$ By Dedekind, this means that there is a 4-cycle $\tau\in G$. The automorphism $\tau$ keeps two of the roots fixed. Because $G$ acts on the set of six roots transitively, we can without loss of generality assume that those two roots are $x_1$ and $x_4=-x_1$ (we may need to replace $\tau$ with its conjugate in $G$ to achieve this).
- Because there are no elements of order four in $S_3$, we see that the restriction $\tau^2$ to $K$ must be the identity. Because $\tau^2(y_i)=y_i, i=1,2,3,$ it follows that
as a product of two disjoint 2-cycles, $\tau^2$ changes the signs of two pairs of roots.
- As $L$ is gotten from $K$ by adjoining some square roots, $H=Gal(L/K)$
must be elementary 2-abelian. Any automorphism $\alpha\in H$ is fully determined by the choice of three signs $\epsilon_i\in\{\pm1\}$, $\alpha(\sqrt{y_i})=\epsilon_i\sqrt{y_i}, i=1,2,3.$. Conjugation by $\sigma$ shifts those three signs cyclically. Furthermore, we have seen $\sigma^3$ has $\epsilon_1=\epsilon_2=\epsilon_3=-1$,
$\tau^2$ has two minus signs, and the conugate $\sigma\tau^2\sigma^{-1}$ has two minus signs at different positions. It follows that the group generated by $\sigma^3$,
$\tau^2$ and $\sigma\tau^2\sigma^{-1}$ gives all the eight sign combinations.
- Therefore $[L:K]=8$ and $Gal(L/K)\simeq C_2\times C_2\times C_2$.
It follows that $[L:\Bbb{Q}]=48=|G|$. Also, we see (a bit of work needed there) that the Galois group is the wreath product $G\simeq C_2\wr S_3$ consisting of all signed permutations of the three pairs of roots.
