Can I use L'Hospitals rule here? (seem like i use it too often...)
Does l'hospitals rule work here?:
[tex]\lim_{h\to_0}\frac{f\left(8+h\right)f\left(8\right)}{h}[/tex] for [tex]f\left(x\right)=x^\frac{4}{3}[/tex] then i would get, [tex]\lim_{h\to_0}\frac{f\left(8+h\right)}{1}[/tex] then, [tex]=\frac{\left(8\right)^\frac{4}{3}}{1}[/tex] ? is that it?... 
No, you can't use it there, in my opinion. The question is asking you to work out the derivative of x^{4/3} at x=8, thus you can't invoke l'Hopital which a priori assumes the function to be differentiable. A minor quibble, but given the way you've written it it seems clear that you're being asked to prove the derivative exists, so assuming it does is not allowed.

i tried factoring according to this:
[tex]a^4b^4=\left(a^2+b^2\right)\left(a+b\right)\left(ab\right)[/tex] but im stuck because i cant get a multiple of h to cancel the denomiator's h. 
Where did the power of 1/3 vanish to? (Note I don't have a simple solution in mind. But it is easy to show that x^4 is differentiable, as is x^3, and hence x^1/3 by the inverse function theorem, hence x^{4/3} is diffible, and it all follows some what simply from these big sounding theorems).

...we haven't touched those theorems yet.
but, [tex]a=\left(8+h\right)^\frac{1}{3}[/tex] and [tex]b=\left(8\right)^\frac{1}{3}[/tex] 
x^n is differentiable at any "x", no matter the value of "n>0". The derivative is equal to n x^(n1). Just plug n=4/3 and x=8 to see what you get.

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