Here is my answer to the question. We start from
FunctionDomain[Surd[x+Sqrt[2-x^2],3]*Surd[1 - x*Sqrt[2 - x^2], 6]/Surd[1 - x^2, 3],x,Reals]
-Sqrt[2] <= x < -1 || -1 < x < 1 || 1 < x <= Sqrt[2]
Therefore, the domain consists of three intervals. On every one of these intervals the function under consideration is constant. This follows from
FullSimplify[D[Surd[x + Sqrt[2 - x^2], 3]*Surd[1 - x*Sqrt[2 - x^2], 6]/Surd[1 - x^2, 3], x]]
Piecewise[{{0, x*Sqrt[2 - x^2] <= 1}}, Indeterminate]
and
Reduce[x Sqrt[2 - x^2] <= 1, x, Reals]
-Sqrt[2] <= x <= Sqrt[2]
and the mean value theorem. The values of these constants can be found by substitutions x==-Sqrt[2] and x==0 and x==Sqrt[2].
Addition. Another approach consists in
Resolve[Exists[c,Resolve[ ForAll[x, x > 1 && x <= Sqrt[2], f == c]]c], Reals]
//ToRadicals
Resolve[Exists[c,
c == -2^(1/6)
Resolve[ ForAll[x, x < -1 && x >= -Sqrt[2], f == c]]c], Reals]//ToRadicals
Both above commands result in True.
c == 2^(1/6)