You need to solve the equation f'(x) = 0 to find the critical values of the function such that:

`f'(x) = (1/3)*x^(1/3 - 1)*((x+3)^(2/3)) + (2/3)x^(1/3)*(x+3)^(2/3-1)`

`f'(x) = (1/3)*x^(-2/3)*((x+3)^(2/3)) + (2/3)x^(1/3)*(x+3)^(-1/3)`

You need to solve the equation f'(x) = 0 such that:

`(1/3)*x^(-2/3)*((x+3)^(2/3)) + (2/3)x^(1/3)*(x+3)^(-1/3) = 0`

`(1/3)x^(1/3)*(x+3)^(2/3)(x^(-1) + 2(x+3)^(-1)) = 0`

`x^(1/3) = 0 =gt x = 0`

`(x+3)^(2/3) = 0 =gt x + 3 = 0=gt x = -3`

`(x^(-1) + 2(x+3)^(-1)) = 0`

`1/x + 2/(x+3) = 0`

`x + 3 + 2x = 0= gt 3x + 3 = 0 =gt 3x =-3 =gt x = -1`

Hence, evaluating the critical values of the function yields x = -3, x = -1 , x = 0.