<s>[INST] find extremum of $y = x(x-1)^{\frac{1}{3}}$ given function:
$$y = x(x-1)^{\frac{1}{3}}$$
steps:
$$y'=\frac{(x^2-x)^{\frac{-2}{3}}2x-1}{3}$$
After simplifying:
$$ y' = \frac{2x-1}{(3x^2-3x)^{\frac{2}{3}}}$$
therefore
$$ x = \frac{1}{2}$$ $$x \ne 0$$
Am I right with calculus?
 [/INST] The function is
$$
f(x)=x\sqrt[3]{1-x}
$$
not
$$
\sqrt[3]{x(1-x)}
$$
at least according to common notation. Thus the derivative is
$$
f'(x)=(1-x)^{1/3}+x\cdot\frac{1}{3}(1-x)^{-2/3}\cdot(-1)=
\frac{1}{3(1-x)^{2/3}}(3-3x-x)=\frac{3-4x}{3(1-x)^{2/3}}
$$
which vanishes at $x=3/4$ and is undefined at $x=1$.
Note that the function is increasing over $(-\infty,3/4]$ and decreasing over $[3/4,\infty)$ (the point where the function is not differentiable doesn't harm).

Alternatively, use the fact that $x\mapsto x^3$ is a continuous bijection $\mathbb{R}\to\mathbb{R}$, so the function $f$ has the same extremal points as
$$
g(x)=(f(x))^3=x^3(1-x)=x^3-x^4
$$
Since $g'(x)=3x^2-4x^3=x^2(3-4x)$, we get the same conclusions as before.
</s>