Study this function $f(x) = \frac{\sqrt[3]{x-1}}{(x+2)^2}$ I need to study this function:
$$f(x) = \frac{\sqrt[3]{x-1}}{(x+2)^2}$$
and I need to show Max and Min point.
The first thing is define the Domain, so:
$$\left\{\begin{matrix}
\sqrt[3]{x-1} > 0\\ 
(x+2)^2 \neq 0
\end{matrix}\right.$$
$$\left\{\begin{matrix}
x > 1\\ 
x \neq -2
\end{matrix}\right.$$
So my domain is:
$$(1, +\infty )$$
Now I check the intersection with x and y:
$$\frac{\sqrt[3]{x-1}}{(x+2)^2} = 0$$
and I get $$ N: x = 1 $$
$$ D: x = -2 $$
And I have no intersection with y.
I have checked the limit in $1$ and $+\infty$:
$$ \lim_{x \rightarrow +\infty} \frac{\sqrt[3]{x-1}}{(x+2)^2} = 0$$
$$ \lim_{x \rightarrow 1^+} \frac{\sqrt[3]{x-1}}{(x+2)^2} = 0$$
$$ \lim_{x \rightarrow 1^-} \frac{\sqrt[3]{x-1}}{(x+2)^2} = 0$$
I have calculated the first derivative as suggested here, the result is:
$$ f'(x) = \frac{8-5x}{3\sqrt[3]{(x-1)^2}(x+2)}$$
Now how should I precede to study Max and Min? Are my steps correct?

First, you must review domain of your function. Why do you need $ \sqrt[3]{x-1} > 0$? Cube root is different from square root, it can have negative value without concerning about complex numbers, as @Vladimir Lenin pointed that out in comments.
Then you can now understand you need to check limits of the given function when $x \to -2$, not only $x \to 1$. Of course $x \to - \infty$ too. Then you'll understand what the function looks like.
Also you need to review your first derivative. The derivative you wrote down is not right.
