Show reciprocal rule for total derivatives $D(\frac{1}{f})(x) = -\frac{1}{f^2(x)} Df(x)$ 
Let $\Omega \subseteq \mathbb{R}^n, f:\Omega \to \mathbb{R}$ differentiable. Show $$D(\frac{1}{f})(x) = -\frac{1}{f^2(x)} Df(x)$$ for all $x \in \Omega$ with $f(x) \neq 0$.

We defined the total derivative as $(V,p)$ and $(W,q)$ finite dimensional vector spaces equipped with the norms p and q, $\Omega \subseteq V$ an open set, $f:\Omega \to W$, $x^* \in \Omega$. $f$ is totally differentiable if there exists a linear map $L:V \to W$ such that $$\lim_{x \to x^*}\frac{f(x) - f(x^*) - L(x - x^*)}{p(x - x^*)} = 0$$ and the linear map $L = Df(x^*)$ is called the differential of $f$ in $x^*$.
We didn't cover the quotient rule for total derivatives yet but I know the chain rule for total derivatives. I don't know how to apply it to the problem, or even apply the definition to the left side of the equation... From the definition it would have to be $\lim_{x \to x*}\frac{f(x) - f(x^*) - D(\frac1f)(x - x^*)}{p(x - x^*)}$ but this makes no sense, I cannot continue from that. Any help is appreciated!
 A: Set
$R(x, x^\ast) = R(x^\ast, x) = \dfrac{1}{f(x)f(x^\ast)} \tag{1}$
and then look at the expression $(f(x^\ast))^{-1} - (f(x))^{-1}$; we have
$\begin{aligned}(f(x^\ast))^{-1} - (f(x))^{-1}&=\dfrac1{f(x^\ast)} - \dfrac{1}{f(x)}\\&=\dfrac{f(x) - f(x^\ast)}{f(x^\ast)f(x)}\\&= R(x, x^\ast)(f(x) - f(x^\ast));\quad (2)\end{aligned}$
noting that
$$\lim_{x \to x^\ast} R(x, x^\ast)  = \dfrac1{(f(x^\ast))^2}  \tag{3}$$
we may use the existence of $Df(x^\ast)$, i.e. that
$$\lim_{x \to x^*}\dfrac{f(x) - f(x^\ast) - Df(x^\ast)(x - x^*)}{p(x - x^*)} = 0 \tag{4}$$
to compute
$\begin{aligned}\lim_{x \to x^*}\dfrac{(f(x))^{-1} - (f(x^\ast))^{-1} + R(x^\ast, x)Df(x^\ast)(x - x^*)}{p(x - x^*)}&= \lim_{x \to x^*}\dfrac{R(x^\ast, x)(f(x^\ast) - f(x)) + R(x^\ast, x)Df(x^\ast)(x - x^*)}{p(x - x^*)}\\&= \lim_{x \to x^*}-R(x^\ast, x)\dfrac{(f(x) - f(x^\ast)) - Df(x^\ast)(x - x^*)}{p(x - x^*)}\\&=\lim_{x \to x^*}-R(x^\ast, x) \lim_{x \to x^*}\dfrac{(f(x) - f(x^\ast)) - Df(x^\ast)(x - x^*)}{p(x - x^*)}\\&=-\dfrac{1}{(f(x^\ast))^2} \times 0 = 0,\quad (5)\end{aligned}$
establishing that
$D(\dfrac{1}{f})(x^\ast) = -R(x^\ast, x^\ast)Df(x^\ast) = -\dfrac{1}{(f(x^\ast))^2} Df(x^\ast). \tag{6}$
QED.
Note Added in Edit:  There's another way to see this, if one has at their disposal the Leibniz Product Rule for derivatives, which says that
$D(fg) = gDf + fDg. \tag{7}$
In the language of ordinary vector calculus, (7) translates to
$\nabla(fg) = f\nabla g + g\nabla f, \tag{8}$
easily verified by simply noting
$\dfrac{\partial (fg)}{\partial x} = f\dfrac{\partial g}{\partial x} + g\dfrac{\partial f}{\partial x} \tag{9}$
for any coordinate variable $x$.  (7) may also be proved under the present, more abstract definition of derivative given by, e.g., (4).  In any event, noting that
$f(x)f^{-1}(x) = 1, \tag{10}$
we have via (7)
$0 = D(1) = D(f(x)f^{-1}(x)) = f^{-1}(x)Df(x) + f(x)Df^{-1}(x), \tag{11}$
which, since $f(x) \ne 0$, may be readily solved for $Df^{-1}(x)$:
$Df^{-1}(x) = -f^{-2}(x)Df(x); \tag{12}$
again I say, QED.
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
A: Hint: Let $g=1/f$, then
$$
g(x)-g(x^*)+g(x^*)^2\cdot L(x-x^*)$$ is $$-g(x)g(x^*)\,(f(x)-f(x^*)-L(x-x^*))-h(x)\,L(x-x^*)$$
where
$$
h(x)=g(x^*)(g(x)-g(x^*))\to0.
$$
Thus,
$$
Dg(x^*)=-g(x^*)^2\cdot L.
$$
A: You can also do as follows:
$$\begin{aligned}\lim_{x\to x}\frac{\left|\frac1{f(x)}-\frac1{f(c)}-\left(-\frac1{f(c)^2}\right)Df(c)(x-c)\right|}{\|x-c\|}&=\lim_{x\to c}\frac{\left|\frac{f(c)-f(x)}{f(x)f(c)}+\frac1{f(c)^2}Df(c)(x-c)\right|}{\|x-c\|}\\&=\lim_{x\to c}\frac{\left|-\frac{f(x)-f(c)-Df(c)(x-c)}{f(x)f(c)}-\frac1{f(x)f(c)}Df(c)(x-c)+\frac1{f(c)^2}Df(c)(x-c)\right|}{\|x-c\|}\\&\le\lim_{x\to c}\frac{\left|-\frac{f(x)-f(c)-Df(c)}{f(x)f(c)}\right|}{\|x-c\|}+\lim_{x\to c}\frac{\left|\frac1{f(c)^2}Df(c)(x-c)-\frac1{f(x)f(c)}Df(c)(x-c)\right|}{\|x-c\|}\\&=\lim_{x\to c}\frac{|f(x)-f(c)-Df(c)(x-c)|}{\|x-c\|}\frac1{|f(x)f(c)|}+\lim_{x\to c}\frac{|Df(x-c)|}{\|x-c\|}\left|\frac1{f(c)^2}-\frac1{f(x)f(c)}\right|\\&\le\lim_{x\to c}\frac{|f(x)-f(c)-Df(c)(x-c)|}{\|x-c\|}\frac1{|f(x)f(c)|}+\lim_{x\to c}\frac{\|Df \|_o\|x-c\|}{\|x-c\|}\left|\frac1{f(c)^2}-\frac1{f(x)f(c)}\right|\\&= \lim_{x\to c}\color{blue}{\frac{|f(x)-f(c)-Df(c)(x-c)|}{\|x-c\|}}\frac1{|f(x)f(c)|}+\lim_{x\to c}\|Df\|_o\color{blue}{\left|\frac1{f(c)^2}-\frac1{f(x)f(c)}\right|}\\&=0.\end{aligned}$$
Now, squeeze.
Note: $\|\cdot\|_o$ denotes the operatornorm and the blue expressions tend to $0$.
