Quotient rule for multivariable functions $\newcommand{\mbf}{\mathbf}$
Let $f,g:\mathbb R^n\to \mathbb R$ be differentiable at $a$.
$Df(a)$ is the unique linear transformation $\mathbb R^n\to\mathbb  R$ such that
$$\lim_{\mbf h\to\mbf 0}\frac{|f(\mbf a+\mbf h)-\mbf f(\mbf a)-Df(\mbf a)(\mbf h)|}{|\mbf h|}=0.$$
I want to show that if $g(a)\neq 0$, then $D(f/g)(a)=\dfrac{g(a)Df(a)-f(a)Dg(a)}{[g(a)]^2}$.
I've shown that $D(f\cdot g)(a)=g(a)Df(a)+f(a)Dg(a)$.
Let $h:\mathbb R^n\to\mathbb R$ be defined by $h(x)=\frac{1}{g(x)}$. Then
$$D(f/g)(a)=D(f\cdot h)(a)=h(a)Df(a)+f(a)Dh(a).$$
Since $h=q\circ g$ where $q:\mathbb R-\{0\}\to\mathbb R$ is defined by $q(x)=1/x$,
$$Dh(a)=D(q\circ g)(a)=Dq(g(a))\circ Dg(a)=D\frac1{g(a)}\circ Dg(a)$$
How can I simplify $D\frac 1{g(x)}$ to prove this?

Just for fun: $Dq(g(a))$ is the derivative of $q : \mathbb R \setminus \{0\} \to \mathbb R$ at the point $g(a) \in \mathbb R \setminus \{0\}$, which can be more helpful writing it as $(Dq)(g(a))$, viewing $Dq$ as a map from $\mathbb R$ to the space of all linear transformations $\mathbb R^n \to \mathbb R$. So, the symbol $D \frac{1}{g(a)}$ doesn't make sense.
Anyway, it is well known that $q'(y) = -y^{-2}$ for any $y \in \mathbb R \setminus \{0\}$, which means that
$$\lim_{h \to 0} \frac{q(y+h)-q(y)+y^{-2}h}{h} = \lim_{h \to 0} \bigg( \frac{q(y+h)-q(y)}{h} + y^{-2} \bigg) = -y^{-2}+y^{-2} = 0$$
and so
$$\lim_{h \to 0} \frac{|q(y+h)-q(y)+y^{-2}h|}{|h|} = 0.$$
Therefore $Dq(y) : \mathbb R \to \mathbb R$ is the map that sends $h$ to $-y^{-2}h$, and in particular, $Dq(g(a))$ is the map that sends $h$ to $-g(a)^{-2}h$, that is, $Dq(g(a))$ is $-g(a)^{-2}$ times the identity map $\mathbb R \to \mathbb R$.
