Answer
$$\frac{d}{d x}\left(\frac{g(x)}{f(x)}\right) =\frac{f(x) g^{\prime}(x)-g(x) f^{\prime}(x)}{f^{2}(x)}$$
Work Step by Step
Since we have $$\frac{d}{d x}\left(\frac{1}{f(x)}\right)=-\frac{f^{\prime}(x)}{f^{2}(x)}\tag{1}$$
Then
\begin{aligned}
\frac{d}{d x}\left(\frac{g(x)}{f(x)}\right)&=g(x)\left(\frac{1}{f(x)}\right)^{\prime}+\frac{1}{f(x)}(g(x))^{\prime},\ \ \text{Use }\ \ (1) \\
&=g(x)\left(\frac{-f^{\prime}(x)}{f^{2}(x)}\right)+\frac{(g(x))^{\prime}}{f(x)}\\
&=\frac{-g(x) f^{\prime}(x)}{f^{2}(x)}+\frac{f(x)\left(g^{\prime}(x)\right)}{f^{2}(x)}\\
& =\frac{f(x) g^{\prime}(x)-g(x) f^{\prime}(x)}{f^{2}(x)}
\end{aligned}
