To get the extrema, derive the function. You get y' = 2x^-1/3 - 2. Set this equal to zero, and you get x=0 as the location of a critical point. Since you are on a closed interval [-1, 1], those points can also have an extrema. Your min is right, but the max isn't at (1,1). At x=-1, you get y=5 (y = 3(-1)^2/3 -2(-1); (-1)^2/3 = 1, not -1). Thus, the maximum is at (-1, 5).
