Let $f:\Bbb R \to\Bbb R^3$ be a differentiable map with $\Vert f(x)\Vert=1$ for any $x \in\Bbb R$. Prove that $f'(x)\cdot f(x)=0.$ Informally I think what I am trying to prove is that at any point on the shell of the sphere of radius 1 centered at (0,0,0) in R3 has a derivative normal to that point's vector.
This is what I have so far:
$\sqrt {f_1(x)^2+f_2(x)^2+f_3(x)^2}=1 \iff f_1(x)^2+f_2(x)^2+f_3(x)^2=1$ 
This is the step that confuses me; am I allowed to just state this?
$\Rightarrow \frac{d}{dx}(f_1(x)^2+f_2(x)^2+f_3(x)^2)=0 \iff \frac{d}{dx}f_1(x)^2+\frac{d}{dx}f_2(x)^2+\frac{d}{dx}f_3(x)^2=0$.
This step is also one I am not sure if I'm allowed to do
$\iff 2(f_1(x)\frac{df_1}{dx}(x)+f_2(x)\frac{df_2}{dx}(x)+f_3(x)\frac{df_3}{dx}(x))=0 \iff f_1(x)\frac{df_1}{dx}(x)+f_2(x)\frac{df_2}{dx}(x)+f_3(x)\frac{df_3}{dx}(x)=0=f(x)\cdot\frac{df}{dx}(x)\Box$
 A: Just take the equality $f_1(x)^2+f_2(x)^2+f_3(x)^2=1$ and differentiate both sides with respect to $x$. What do you get?
The two steps you are confused about are linearity of differentiation: $\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx} f(x)+\frac{d}{dx} g(x)$ and $\frac{d}{dx}(af(x))=a\frac{d}{dx}f(x)$ for functions $f$ and $g$ and a real number $a$. 
A: The fact $\lVert f(x)\rVert=1$ implies $$\lVert f(x)\rVert^2=\langle f(x),f(x)\rangle=1$$ 
Now take the derivative of both sides with respect to $x$ to get 
$$2\langle f(x),f'(x)\rangle=0$$
The meaning of that is, that the position vector is perpendicular to the velocity vector.
Edit: 
The second part is easy to obtain
$$\frac{d}{dx}\langle f(x),f(x)\rangle=\langle f'(x),f(x)\rangle+\langle f(x),f'(x)\rangle=2\langle f'(x),f(x)\rangle=2\langle f(x),f'(x)\rangle$$
by product rule and symmetry.
A: Since
$(f_1(x))^2 + (f_2(x))^2 + (f_3(x))^2 =1, \tag 1$
differentiating we have
$2f_1(x) f_1'(x) + 2f_2(x) f_2'(x) +2f_3(x) f_3'(x) = 0; \tag 2$
we divide by $2$ and obtain
$f(x) \cdot f'(x) = f_1(x) f_1'(x) + f_2(x) f_2'(x) +f_3(x) f_3'(x) = 0. 
 \tag 3$
Your demonstration agrees with mine, hence it must be correct! 
