How to rationalize multiple terms with fractional exponents

The numerator is a difference of cube-roots, not a difference of cubes. So rationalizing the numerator: $$ \frac{(x+h)^{2/3}-x^{2/3}}{h}= \frac{(x+h)^2-x^2}{h[(x+h)^{4/3}+x^{4/3}(x+h)^{2/3}+x^{4/3}]} $$ gives the factor of $h$ you want in the numerator. Hence \begin{align*} \require{cancel} \lim_{h\to 0} \frac{(x+h)^{2/3}-x^{2/3}}{h} &=\lim_{h\to 0} \frac{\cancel{h}(2x+h)}{\cancel{h}[(x+h)^{4/3}+x^{4/3}(x+h)^{2/3}+x^{4/3}]}\\ &=\frac{2x}{3x^{4/3}}=\frac23 x^{-1/3}. \end{align*}

In general, $A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\dots+B^{n-1})$, so $$ \alpha^{m/n}-\beta^{m/n}=\frac{\alpha^m-\beta^m}{\alpha^{(n-1)m/n}+\alpha^{(n-2)m/n}\beta^{m/n}+\dots+\beta^{(n-1)m/n}} $$ which allows you to deal with the derivative of $x^{m/n}$.

Idea

You have $$\frac{A-B}{h}=\frac{(A-B)}{h}\frac{(A^2+AB+B^2)}{(A^2+AB+B^2)}=\frac{A^3-B^3}{h(A^2+AB+B^2)}.$$ Now let $A=(x+h)^{2/3}$ and $B=x^{2/3}$