[(x+h)^1/3 - x^1/3] / h

Can you show the steps to get this to (x^-2/3)/3?

Let (x+h)^1/3=a -and x^1/3=b. So a^3-b^3= h
(a-b)/h=(a-b)(a^2+ab+b^2)/(a^2+ab+b^2)h
(a^3 - b^3)/(a^2+ab+b^2)h
h/(a^2+ab+b^2)h
1/(a^2+ab+b^2)
1/([(x+h)^2/3 +(x+h)^1/3 x^1/3+ x^2/3}
For h close to zero
1/{3 x^2/3}=(x^-2/3)/3

Limit of numerator = 0, limit of denominator = 0 => we can use L'Hôpital's rule.
Differentiate the numerator w.r.t h and we get ((x+h)^-2/3)/3 and differentiate the denominator and we get
1. The limit of this is (x^-2/3)/3 (setting h=0)

hmm......
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jeez........ sorry, i don't know