Side limits of the derivative of this function $f:\mathbb{R}\to \mathbb{R}$ with $f\left(x\right)=\left(x^3+3x^2-4\right)^{\frac{1}{3}}$
Calculate side limits of this function's derivative, $f'_s\:and\:f'_d$, in $x_o=-2$
The answer key says I should get $\infty $ and $-\infty$ but I'm not getting that. The derivative I get is $\frac{x\left(x+2\right)}{\left(\left(x-2\right)^2\left(x+2\right)^4\right)^{\frac{1}{3}}}$ and by doing the multiplication from the denominator I would get something with $x^2$.

You have the wrong expression for the derivative.
$$\begin{align}
\frac d{dx}\left[(x^3+3x^2-4)^{1/3}\right]
  &= \frac 13(x^3+3x^2-4)^{-2/3}(3x^2+6x) \\[2ex]
  &= \frac {x^2+2x}{[(x+2)^2(x-1)]^{2/3}} \\[2ex]
  &= \frac {x(x+2)}{(x+2)^{4/3}(x-1)^{2/3}} \\[2ex]
  &= \frac {x}{(x+2)^{1/3}(x-1)^{2/3}} \\[2ex]
\end{align}$$
That last expression's denominator tends to zero as $x\to-2$ but the numerator does not tend to zero, which means an infinite limit on both sides of $-2$. As $x\to-2$ from the left, both numerator and denominator are negative, so the expression tends to $+\infty$. As $x\to-2$ from the right, the numerator is negative but the denominator is positive, so the expression tends to $-\infty$.
