We could solve the set in the following way:
Case I: Solve for x^^2 < 2x and 2x < 1/x.
You would get x < 2 and x < 1/sqrt(2).
So, x < 1/sqrt(2) satisfies the set.
Case II: Solve for x^^2 < 1/x and 1/x < 2x.
You would get x^^3 < 1 and 2x^^2 > 1
So, for all real x, x <1 and x > 1/sqrt(2).
So, 1/sqrt(2) < x < 1 satsifies the set.
Case III: Solve for 2x < x^^2 and x^^2 <1/x.
You would get x >2 and x^^3 < 1.
For all real x, you must have x < 1 (from the second equation).
Now x < 1 and x >2 are incompatible.
So III does not hold good for all real x.
So I & II are correct.
Be cautious not to select values such as x =1, x = 2.
For these values, some of the functions yield the same value e.g. for x =1, x^^2 = 1/x = 1, for x = 2, 2x = x^^2 = 4.
So you would not get the desired relationships.