Answer
See graph and explanations.
Work Step by Step
See graph; when $x\to0^+, y\to\infty$ or $\lim_{x\to 0^+}(x^{2/3}+\frac{1}{x^{1/3}})=\infty$;
while when $x\to0^-, y\to-\infty$ or $\lim_{x\to 0^-}(x^{2/3}+\frac{1}{x^{1/3}})=-\infty$;
the asymptote of the function is $x=0$
