Using the definition of the derivative prove that if $f(x)=x^\frac{4}{3}$ then $f'(x)=\frac{4x^\frac{1}{3}}{3}$ So I have that $f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ 
and know that applying  
$f(x)=x^\frac{4}{3}=f\frac{(x+h)^\frac{4}{3} -x^\frac{4}{3}}{h}$
but am at a loss when trying to expand 
$(x+h)^\frac{4}{3}$

You can use the generalized binomial coeficients to solve this problem, which gives you
$(x + h)^\frac{4}{3} = x^\frac{4}{3}(1 + \frac{4h}{3x} + \frac{4h^2}{18x^2} + \ldots)$
It will go on forever, but since you divide by $h$, most will tend to zero, and you'll be left with $x^\frac{4}{3} \frac{4}{3x} = \frac{4}{3}x^\frac{1}{3}$.
