Locate the absolute extrema of the function on the closed interval:
y = 3x^(2/3) - 2x, [-1, 1]

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).

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