You have already been given two nice ways to solve the problem, so it is time for an ugly way.
Bring to a common denominator. Multiply top and bottom by $(n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3}$ and use the identity $x^3-y^3=(x-y)(x^2+xy+y^2)$ with $x=(n+1)^{1/3}$ and $y=n^{1/3}$. Our expression simplifies to $\frac{2}{n^{1/3}(n+1)^{1/3}\left( (n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3}\right)}.$ This is $\lt \dfrac{1}{n^{4/3}}$.
Another way: The following is better, no "magic" identities. Let $f(x)=\dfrac{2}{x^{1/3}}$. Then $f'(x)=-\dfrac{2}{3x^{4/3}}$. By the Mean Value Theorem, there is a $c$ between $n$ and $n+1$ such that $f(n+1)-f(n)=-\frac{2}{3c^{4/3}}.$ It follows that our $n$-th term, which is $f(n)-f(n+1)$, is $\lt \dfrac{2}{3n^{4/3}}$. We conclude convergence from the Comparison Test.