If $f: U \rightarrow \mathbb R^n$ ($U \subset \mathbb R^m$ is an open set) is differentiable and $f(x) \neq 0$ $\forall x \in U$ $\Rightarrow$ $\varphi: U \rightarrow \mathbb R$, $\varphi(x) = \frac {1}{||f(x)||}$ is differentiable.
I know how to show that $\varphi$ is differentiable, but I'm having problems to find the differential $\varphi'(x).v$, $\forall v \in \mathbb R^m$.
Consider first $\phi(x)=\|f(x)\|^2=\sum_i f_i(x)^2$ so that $\frac{\partial \phi}{\partial x_j}=2 \sum_i f_i(x) \frac{\partial f_i}{\partial x_j}(x)$. Now, observe $\varphi(x)=\frac{1}{\sqrt{\phi(x)}}$ and apply the chain rule: $$\frac{\partial \varphi}{\partial x_j}=\frac{d \varphi}{d \phi}\frac{\partial \phi}{\partial x_j}=-\frac{1}{2}\frac{1}{(\phi(x)^{\frac{3}{2}}}\frac{\partial \phi}{\partial x_j}=-\frac{1}{\|f(x)\|^3} \sum_i f_i(x) \frac{\partial f_i}{\partial x_j}(x)$$
One has (where it makes sense) $$\varphi '(x)=-\frac{1}{\|f(x)\|^2}\left(x\mapsto\|f(x)\|\right)'.$$ Now, as $\dim\mathbb{R}^n=n<+\infty$, if $f(x)=\left(f_1(x),\ldots,f_n(x)\right)$, we can write $$\|f(x)\|=\sqrt{f_1(x)^2+\ldots+f_n(x)^2}$$ and then $$\frac{\partial}{\partial x_i}\left(x\mapsto\|f(x)\|\right)=\frac{2f_i(x)f_i'(x)}{2\sqrt{f_1(x)^2+\ldots+f_n(x)^2}}\quad\quad\quad 1\leq i \leq n$$ so $$\left(x\mapsto\|f(x)\|\right)'=\frac{1}{\sqrt{f_1(x)^2+\ldots+f_n(x)^2}}\sum_{i=1}^{n}f_i(x)f'_i(x)\mathrm{d}x_i\in\left(\mathbb{R}^n\right)^*.$$
Hence, we get $$\varphi '(x)\left(\left(h_1,\ldots,h_n\right)\right)=-\frac{1}{\left(f_1(x)^2+\ldots+f_n(x)^2\right)^{3/2}}\sum_{i=1}^{n}f_i(x)f'_i(x)\mathrm{d}x_i.$$
Name $$\begin{array}{l|rcl} g : & \mathbb R^n & \longrightarrow & \mathbb R \\ & x & \longmapsto & \Vert x \Vert^2 =\langle x,x \rangle \end{array}$$
$g$ is differentiable and $g^\prime(x).h = 2 \langle x,h \rangle$.
Also name $$\begin{array}{l|rcl} h : & \mathbb R^* & \longrightarrow & \mathbb R \\ & x & \longmapsto & \frac{1}{\sqrt{x}} \end{array}$$
$h$ is differentiable and $h^\prime(x)=-\frac{1}{2x^{3/2}}$.
Now you can notice that $$\varphi(x)=(h \circ g \circ f)(x)$$ You can apply the chain rule to find $$\varphi^\prime(x).v=-\frac{\langle f(x),f^\prime(x).v\rangle }{\Vert f(x) \Vert^3}$$ The formula is valid for any Hilbert space, i.e. of finite or infinite dimension.
This is a chain rule problem, where $\varphi =g\circ f$, with $g=(\left \| \cdot \right \|)^{-1}$.
$f:\mathbb R^{m}\to \mathbb R^{n}$
$g:\mathbb R^{n}\to \mathbb R$ so that
$g\circ f:\mathbb R^{m}\to \mathbb R$
and
$\varphi '(x)=g'(f(x))\circ f'(x)$
Now, $f'(x)=\textbf D_{f}(x)$ and $g'(f(x))=(\nabla g)(f(x))$ so
$g'(f(x))(\vec h)=(\nabla g)(f(x))(\vec h)=\frac{-f(x)}{\left \| f(x) \right \|^{3/2}}(f_{1}(x),\cdots ,f_{n}(x))\cdot (h_{1},\cdots ,h_{m})$ so
$(\textbf D_{f}(x)\circ(\nabla g)(f(x))(\vec h) =$
$\frac{-f(x)}{\left \| f(x) \right \|^{3/2}}\left ( \sum_{i=1}^{n}f_{i}(x)\frac{\partial f_{1}}{\partial x_{i}}(x),\cdots ,\sum_{i=1}^{n}f_{i}(x)\frac{\partial f_{m}}{\partial x_{i}}(x) \right )\cdot (h_{1},\cdots ,h_{m})$
Short (maybe cryptic) answer: $$ \nabla \varphi (x) = \nabla \left(\frac{1}{\|f(x)\|}\right) = -\frac{ \nabla (\|f(x)\|)}{\|f(x)\|^2} = - \frac{\nabla f(x) \, f(x)}{\|f(x)\|^3}. $$
