$\textbf{My problem:}$ Let $f:\mathbb{R^m}\rightarrow \mathbb{R^m}$ be a function of class $C^1$ such that for each $x\in \mathbb{R^m}$, $|f'(x)\cdot v|=|v|$. Prove that $|f(x)-f(y)|=|x-y|$ for each $x, y\in \mathbb{R^m}$.
$\textbf{My attempt:}$ I take $a,b\in \mathbb{R^m}$ distincts. then I consider the function $\varphi :[0,1]\rightarrow \mathbb{R}$ defined by $\varphi (t)=\langle f(a(1-t)+bt),b-a\rangle$. I have that $\varphi $ is continuous on $[0,1]$ and differentiable on $(0,1)$. Then for mean value theorem, there exist $x\in (0,1)$ such that $\varphi (1)-\varphi (0)=\varphi '(x)$. Whit this, I can conclude easily:
$\langle f(b)-f(a)-f'(a+x(b-a))\cdot (b-a), b-a\rangle=0$
I want deduce from this that $f(b)-f(a)-f'(a+x(b-a))\cdot (b-a)=0$, because in this case, I'd have $f(b)-f(a)=f'(a+x(b-a))\cdot (b-a)$, which implies $|f(b)-f(a)|=|f'(a+x(b-a))\cdot (b-a)|=|b-a|.$ However, I don't achieve to prove that $f(b)-f(a)-f'(a+x(b-a))\cdot (b-a)=0$...Is impossible deduce this fact from my develop?...any suggestions?