The first thing you should note is that the expression
is undefined at any zero of the denominator.
So, the points $x=\pm \sqrt 2$, $\pm 2$, and $\pm\sqrt 6$ are
not in the domain of $\sqrt\Phi$.
Now, to find the domain of $\sqrt\Phi$, we need to find when $\Phi$ is nonnegative.
Towards solving $\Phi\ge0$,
the fundamental observation to make is that
the only points "across which" $\Phi$ can change signs are at the zeros of its numerator or at the zeroes of its denominator.
The zeroes of the numerator are: $$\pm 1, \pm \sqrt 3, \pm \sqrt 5$$
and the zeroes of the denominator are
\pm \sqrt 2, \pm 2 \pm\sqrt 6.
As already mentioned, the only points across which $\Phi$ can change sign are at one of the zeroes above.
Let's also note the expression $\Phi$ is "even": if $\Phi(x)\ge 0$, then $\Phi(-x)\ge 0$. So, let's find the $x$ values on the nonnegative $x$-axis that satisfy $\Phi(x)\ge0$. Then by "reflection" we'll obtain the points on the negative $x$-axis where $\Phi(x)\ge0$.
So, draw a number line with the zeroes listed above:
The zeroes subdivide the nonnegative $x$-axis into intervals, the endpoints of which, except the endpoint 0, are the zeroes of either the numerator or the denominator of $\Phi$. In each subinterval, if you pick a point $x_0$ (not an endpoint, save for 0), evaluate $\Phi(x_0)$, and note its sign, then across
that entire subinterval, the expression $\Phi$ will have that sign.
For example in $[0,1)$, picking $x=1/2$ it is easy to see that the sign of $\Phi(1/2)$ is
So on all of $[0,1)$, the expression $\Phi$ is positive. (Note, you only need to find the sign, there is no need to do the actual arithmetic.)
The complete "sign chart" is shown below:
And we see that $\Phi>0$ on
By symmetry, $\Phi>0$ on
( -\infty,-\sqrt6)\cup(-\sqrt5,2)\cup (-\sqrt3,-\sqrt2)\cup (-1,0].
So, the domain of $\sqrt\Phi$ is the union of the two sets above, together with
the zeroes of the numerator of $\Phi$ that aren't zeros of the denominator of $\Phi$: $x=\pm1$, $x=\pm \sqrt3$, and $x=\pm \sqrt5$.
Below is a plot of $y=\Phi$ and $y=\sqrt\Phi$. Note that the zeroes of the denominator of $\Phi$ are vertical asymptotes of the graph of $y=\Phi$ and the zeroes of the numerator of $\Phi$ are the zeroes of $\Phi$.