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 $$f'(x) = \dfrac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$$

I think that a better answer to this question is the formula \begin{equation}\label{Sitnik-Bourbaki}\tag{1} \frac{\textrm{d}^k}{\textrm{d}z^k}\biggl(\frac{u}{v}\biggr) =\frac{(-1)^k}{v^{k+1}} \begin{vmatrix} u & v & 0 & \dotsm & 0\\ u' & v' & v & \dotsm & 0\\ u'' & v'' & 2v' & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \ddots & \dotsm\\ u^{(k-1)} & v^{(k-1)} & \binom{k-1}1v^{(k-2)} & \dots & v\\ u^{(k)} & v^{(k)} & \binom{k}1v^{(k-1)} & \dots & \binom{k}{k-1}v' \end{vmatrix} \end{equation} where $u=u(z)$ and $v=v(z)\ne0$ are differentiable functions. For more information on the formula \eqref{Sitnik-Bourbaki}, please refer to another answer of mine at the site .