When x is negative x2/3 =1 will have three possible solutions, two of them complex. If we are to stick with the reals, then we must do the squaring before the cube rooting. i.e. we must express x2/3 as (x2)1/3.
When x approaches -1 from above then (x2)1/3 -1 is negative, x - 1 is negative, so f(x) is positive, and approaches the limit +infinity.
When x approaches -1 from below then (x2)1/3 -1 is positive, x - 1 is negative, so f(x) is negative, and approaches the limit -infinity.
For x = +1, use l'Hopital's rule to show the limit is 3/2.