We will use the following formula 

`x^3-y^3=(x-y)(x^2+xy+y^2)`                                (1)

By definition

`f'(x)=lim_(Delta x->0)(f(x+Delta x)-f(x))/(Delta x)`

So for `f(x)=root(3)(x)` we have

`f'(x)=lim_(Delta x->0)(root(3)(x+Delta x)-root(3)(x))/(Delta x)`

Now we multiply both numerator and denominator by `(x+Delta x)^(2/3)+(x(x+Delta x))^(1/3)+x^(2/3)` and then we use formula (1) (reading formula from right to left) on numerator to get

`lim_(Delta x->0)(x+Delta x-x)/(Delta x((x+Delta x)^(2/3)+(x^2+x Delta x)^(1/3)+x^(2/3)))=`

`lim_(Delta x->0)(Delta x)/(Delta x((x+Delta x)^(2/3)+(x^2+x Delta x)^(1/3)+x^(2/3)))=`

`lim_(Delta x->0)1/((x+Delta x)^(2/3)+(x^2+Delta x)^(1/3)+x^(2/3))=1/(x^(2/3)+x^(2/3)+x^(2/3))=1/(3x^(2/3))=`

`1/(3root(3)(x^2))`

So your solution is `f'(x)=1/(3root(3)(x^2))`