Is this method okay to show that $f(x)=x^{-2/3}$ is discontinuous at point $x= 0$? Limit $f(x)$ as we tends from positive side of $0$, we get $+\infty$ as the value, same with $f(x)$ approached from left side, there we also get $f(x)$ to be $+ \infty$, but as $x$ is not in domain of the function we may say that $f(x)$ is discontinuous at $x=0$?
If we happen to define $f(x)$ at zero to be $+\infty$, then what can be said about continuity at $x= 0$? Or we will argue that limit value which is infinity, we cannot surely say that it is the same value as the $f(x)$ defined value of infinity, they are all non-properly defined, hence again we have a discontinuous nature?
On the domain $\mathbb{R} \setminus \{0\} = (-\infty, 0) \cup (0, \infty)$, the function $x \mapsto x^{-2/3}$ is indeed continuous, assuming that you interpret that expression as $$ x^{-2/3} = \bigl(x^2\bigr)^{-1/3} = \bigl\lvert x^2 \bigr\rvert^{-1/3} = \lvert x \rvert^{-2/3}. $$ Also, your observation that the limit as $x \to 0$ is $+\infty$ is correct. However, that's just shorthand notation for saying that "the values of the function $f$ increase without bound as $x$ approaches $0$." Strictly speaking that means that the limit does not exist. So there's no way to define $f(0)$ in order to make a function that's continuous at $0$ that agrees with the given formula.