Your first derivative looks good (although, I think you have a typo: the last factor should be $(x-1)^2$, not $(x-2)^2$). After cleaning it up, this is what I get:

(1)
\begin{align} f'(x)=\frac{8(x^2+x-2)}{3(x+3)^{1/3}}=\frac{8(x-1)(x+2)}{3(x+3)^{1/3}} \end{align}

This has the same critical values that you obtained. For the second derivative, I get (before simplifying):

(2)
\begin{align} f''(x)=\frac{8(2x+1)3(x+3)^{1/3}-8(x^2+x-2)(x+3)^{-2/3}}{9(x+3)^{2/3}} \end{align}

Now, because of the negative exponent in the second term in the numerator, you need to a little house cleaning. As an intermediate step, you should encounter:

(3)
\begin{align} f''(x)=\left(24(2x+1)(x+3)^{1/3}-\frac{8(x^2+x-2)}{(x+3)^{2/3}}\right) \cdot \frac{1}{9(x+3)^{2/3}} \end{align}

This becomes

(4)
\begin{align} f''(x)=\frac{8(5x^2+20x+11)}{9(x+3)^{4/3}} \end{align}

The bottom yields a critical value of $x=-3$ and the numerator has two more critical values that you can obtain with the quadratic formula.

This problem is pretty epic, huh?