quotient rule

\begin{align*} \left( \frac{f(x)}{g(x)} \right)^{\prime}&=\lim_{h\to0}\frac{\displaystyle \frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}\\ &=\lim_{h\to0}\left( \frac{g(x)f(x+h)-f(x)g(x+h)}{h}\cdot\frac{1}{g(x+h)g(x)} \right)\\ &=\lim_{h\to0}\frac{g(x)f(x+h)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{h}\cdot \frac{1}{g(x)g(x)}\\ &=\lim_{h\to0}\left( g(x)\frac{f(x+h)-f(x)}{h}-f(x)\frac{g(x+h)-g(x)}{h} \right)\cdot \frac{1}{(g(x))^{2}}\\ &=\left( g(x)f^{\prime}(x)-f(x)g^{\prime}(x) \right)\cdot \frac{1}{(g(x))^{2}}\\ &=\frac{g(x)f^{\prime}(x)-f(x)g^{\prime}(x)}{(g(x))^{2}\\} \end{align*}

A simpler proof can be done with Leibniz product rule and reciprocal rule.
\[\frac{f(x)}{g(x)}=f(x)\cdot \frac{1}{g(x)}\]

Using the product rule, \[f(x)\cdot \frac{1}{g(x)}=f^{\prime}(x)\cdot \frac{1}{g(x)}+f(x)\cdot \left( \frac{1}{g(x)} \right)^{\prime}\]. Now with reciprocal rule, \[\left( \frac{1}{g(x)} \right)^{\prime}=-\frac{g^{\prime}(x)}{(g(x))^{2}}\]. Combining both of them:

\begin{align*} f(x)\cdot \frac{1}{g(x)}&=f^{\prime}(x)\cdot \frac{1}{g(x)}+f(x)\cdot -\frac{g^{\prime}(x)}{(g(x))^{2}}\\ &=\frac{f^{\prime}(x)}{g(x)}-\frac{f(x)g^{\prime}(x)}{(g(x))^{2}}\\ &=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{(g(x))^{2}}\\ \end{align*}