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?

You only need to show that $D\frac{1}{g(a)}=-\frac{Dg(a)}{g^2(a)}$ and for this is necessary and sufficient that
$$\lim_{h\to 0}\frac{|1/g(a+h)- 1/g(a)+D(g(a))(h)/g^2(a)|}{|h|}=0.$$
Indeed, note that
$$\frac{|1/g(a+h)- 1/g(a)+D(g(a))(h)/g^2(a)|}{|h|}=\frac{|g^2(a)-g(a)g(a+h)+g(a+h)Dg(a)(h)|}{|g^2(a)g(a+h)h|}\leq\frac{|g^2(a)-g(a)g(a+h)+g(a)Dg(a)(h)|}{|g^2(a)g(a+h)h|}+\frac{|g(a+h)Dg(a)(h)-g(a)Dg(a)(h)}{|g^2(a)g(a+h)h|}=\frac{|g(a+h)-g(a)-Dg(a)(h)|}{|g(a)g(a+h)||h|}+\frac{|(g(a+h)-g(a))Dg(a)(h/|h|)}{|g^2(a)g(a+h)|},$$
where making $h\to0$ the first summand $\to0$ by the differentiability of $g$ and the second summand $\to0$ because it is the product of a function wich converges to $0$ and a bounded function.
