Y = (1 / 3) ^ (2 / 3x-1), find the range of function y=(1/3)^[2/(3x-1)]

Y = (1 / 3) ^ (2 / 3x-1), find the range of function y=(1/3)^[2/(3x-1)]

Because the value range of the function at exponential position is (- ∞, 0) (0, + ∞)
So the range of the original function is (0,1) (1, + ∞)