Find points of discontinuity of $x(x-1)^{2/3}e^{\sqrt[3]{x}}$

Question

I want to find the points of discontinuity for the following function:

$$f(x)=x(x-1)^{2/3}e^{\sqrt[3]{x}}$$

My textbook says it can be derived for every number except $0$ and $1$ which could be points of discontinuity. I have no idea how it reached that conclusion without deriving the function first. Now, I find the first derivative by using the product rule:

$$f'(x)=e^{\sqrt[3]{x}}\left[{\frac {x^{4/3}-x^{1/3}+5x-3}{3(x-1)^{1/3}}}\right]$$

My textbook asks for the domain. I know that $\sqrt[3]{x}$ are defined for any $x$ so I focus on the denominator which should not be equal to $0$, therefore $3(x-1)^{1/3}\neq0\implies x-1\neq0\implies x \neq 1$. Wolfram Alpha says the domain should be $x>1$.

Any hints on what I'm doing wrong?

Answer 1

Probability Wolfram assumes $(x-1)^{2/3}$ defined only for $x-1\ge 0\implies x\ge 1$.

Answer 2

Probably Wolfram assumes $(x−1)^{2/3}$ has abranch cut in $x=1$