The square root exists for [tex]2x+8\ge0\implies x\ge-4[/tex], or on the interval [tex][-4,\infty)[/tex].
But [tex]f(x)[/tex] is undefined when its denominator is zero, which happens for
[tex]x^2-2x=x(x-2)=0\implies x=0\text{ or }x=2[/tex]
so we need to remove these points from the interval above. Then the domain of [tex]f[/tex] is
[tex][-4,0)\cup(0,2)\cup(2,\infty)[/tex]
