The answer is:
$\frac{d^n}{dx^n} \left (\frac{f(x)}{g(x)} \right ) = \sum_{k=0}^n {(-1)^k \tbinom{n}{k} \frac{d^{n-k}\left(f(x)\right)}{dx^{n-k}}}\frac{A_k}{g_{(x)}^{k+1}} $
where:
$A_0=1$
$A_n=n\frac{d\left(g(x)\right)}{dx}\ A_{n-1}-g(x)\frac{d\left(A_{n-1}\right)}{dx}$
for example let $n=3$:
$\frac{d^3}{dx^3} \left (\frac{f(x)}{g(x)} \right ) =\frac{1}{g(x)} \frac{d^3\left(f(x)\right)}{dx^3}-\frac{3}{g^2(x)}\frac{d^2\left(f(x)\right)}{dx^2}\left[\frac{d\left(g(x)\right)}{d{x}}\right] + \frac{3}{g^3(x)}\frac{d\left(f(x)\right)}{d{x}}\left[2\left(\frac{d\left(g(x)\right)}{d{x}}\right)^2-g(x)\frac{d^2\left(g(x)\right)}{dx^2}\right]-\frac{f(x)}{g^4(x)}\left[6\left(\frac{d\left(g(x)\right)}{d{x}}\right)^3-6g(x)\frac{d\left(g(x)\right)}{d{x}}\frac{d^2\left(g(x)\right)}{dx^2}+g^2(x)\frac{d^3\left(g(x)\right)}{dx^3}\right]$
Relation with Faa' di Bruno coefficents:
The $A_n$ have also a combinatorial form, similar to the Faa' di Bruno coefficents (ref http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno).
An explication via an example (with for shortness
$g'=\frac{d\left(g(x)\right)}{dx}$, $g''=\frac{d^2\left(g(x)\right)}{dx^2}$, etc.):
Let we want to find $A_4$.
The partitions of 4 are: $1+1+1+1, 1+1+2, 1+3, 4, 2+2$.
Now for each partition we can use the following pattern:
$1+1+1+1 \leftrightarrow C_1g'g'g'g'=C_1\left(g'\right)^4$
$1+1+2+0 \leftrightarrow C_2g'g'g''g=C_2g\left(g'\right)^2g''$
$1+3+0+0 \leftrightarrow C_3g'g'''gg=C_3\left(g\right)^2g'g'''$
$4+0+0+0 \leftrightarrow C_4g''''ggg=C_4\left(g\right)^3g''''$
$2+2+0+0 \leftrightarrow C_5g''g''gg=C_5\left(g\right)^2\left(g''\right)^2$
with $C_i=(-1)^{(4-t)}\frac{4!t!}{m_1!\,m_2!\,m_3!\,\cdots 1!^{m_1}\,2!^{m_2}\,3!^{m_3}\,\cdots}$ (ref. closed-form of the Faà di Bruno coefficents)
where $t$ is the numers of partition items different of $0$, and $m_i$ is the numer of i.
We have $C_1=24$ (with $m_1=4, t=4$), $C_2=-36$ (with $m_1=2, m_2=1, t=3$), $C_3=8$ (with $m_1=1, m_3=1, t=2$), $C_4=-1$ (with $m_4=2, t=1$), $C_5=6$ (with $m_2=2,t=2$).
Finally $A_4$ is the sum of the formula found for each partition, i.e.
$A_4=24\left(g'\right)^4-36g\left(g'\right)^2g''+8\left(g\right)^2g'g'''-\left(g\right)^3g''''+6\left(g\right)^2\left(g''\right)^2$
Best Answer
Recall that $$f(x) = \dfrac{g(x)}{h(x)} = g(x) \cdot (h(x))^{-1}$$ i.e., invert the denominator of a quotient of functions, after which you can use the product rule. And the chain rule applies, as usual. $$f'(x) = g'(x)[h(x)]^{-1} + g(x)\Big(-[h(x)]^{-2}\cdot h'(x)\Big)$$
Now, simplify (finding common denominator), and you'll have $$f'(x) = \dfrac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$$