In the book of The Elements of Real Analysis by Bartle, at page 247, it is asked that
If $f$ is differentiable on an open subset $D \subseteq \mathbb{R}^p$ and has values in $\mathbb{R}^q$ such that $|f(x)| = 1$ for $x\in D$, then $$f(x) \cdot Df(x)(u) = 0 \quad for \quad x\in D, u \in \mathbb{R}^p.$$ Note: in Bartle's notation $Df(x)(u) = J_f(x) * u$, where $J$ is the jacobin of $f$ at $x \in D$.
I have tried the following:
$$f(x) \cdot [J_f (x) * u] = f(x) \cdot [\sum_{k = 1}^p J_{i, k} u_k]_{i,1} = \sum_{l = 1, k=1}^p f_l(x) J_{l, k} u_k \\ = \sum_{l = 1}^p f_l(x) [\nabla f_l(x) \cdot u]$$ but stuck in here, so my questions are that
i-) How can we continue from the point where I left ?
ii-) How can we show the desired result alternatively ?
Edit:
Further answers are also welcomed, since there might be more intuitive diverse solutions to this question.
Use the chain rule.
Let $g:\newcommand{\RR}{\mathbb{R}}\RR^p\to\RR$ be $g(x) = \|x\|$. Note that $g$ is differentiable away from the origin, with derivative $$D_xg(u)= \sum_{i=1}^p \frac{x_iu_i}{g(x)}=\frac{x\cdot u}{g(x)}.$$ If $\|f(x)\|=1$, then $$D_x (g\circ f)(u) = (D_{f(x)}g\circ D_x f)(u)= 0,$$ but then, noting that $g(f(x))=1$ by assumption, $$0=(D_{f(x)}g) (D_xf(u))= \frac{f(x)\cdot D_xf(u)}{g(f(x))}=f(x)\cdot D_xf(u),$$ as desired.
Also sorry, out of habit, I've used the notation $D_xf := Df(x)$