These used to reaaaally confuse me! So, I think you mean the cubed root of x2 -1.
3√(x2 - 1)
We want two functions so that if we put g(x) into f(x) for x, we get h(x), right? These can get confusing, buuuuut if we pull out what's under the radical sign, it can be g(x), because it's the equation we have that involves x.
So, g(x) = x2 - 1
If we do the reverse of f(g(x)) and replace x2 -1 with x, we end up with 3√x, so f(x) = 3√x.
Let's show it. If we take g(x) = x2 - 1 and place it into f(x) = 3√x for x, we get your given h(x) = 3√(x2 - 1), right?
By the way, if you ever need to throw the cubed root of something in your cslculator, remember that it's the same as the 1/3 power. Another way of writing h(x) is (x2 - 1)1/3.
Hope this helps!