Find the derivative using the chain rule and the quotient rule 
$$f(x) = \left(\frac{x}{x+1}\right)^4$$ Find $f'(x)$.

Here is my work:
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4\left(x+1\right)^3x^4}{\left(x+1\right)^8}$$
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4x^4\left(x+1\right)^3}{\left(x+1\right)^8}$$
I know the final simplified answer to be:
$${4x^3\over (x+1)^5}$$
How do I get to the final answer from my last step? Or have I done something wrong in my own work?

Not wrong, only written in a different form: you can simplify much of it.
It's much better exploiting the chain rule: if you call
$$
g(x)=\frac{x}{x+1},
$$
then $f(x)=(g(x))^4$ and so
$$
f'(x)=4(g(x))^3g'(x)=
4\left(\frac{x}{x+1}\right)^{\!3}\frac{1(x+1)-x\cdot 1}{(x+1)^2}=
\frac{4x^3}{(x+1)^5}
$$
