Vtyjkyuk

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?

$Rightarrow fracddx(f_1(x)^2+f_2(x)^2+f_3(x)^2)=0 iff fracddxf_1(x)^2+fracddxf_2(x)^2+fracddxf_3(x)^2=0$.

This step is also one I am not sure if I'm allowed to do

$iff 2(f_1(x)fracdf_1dx(x)+f_2(x)fracdf_2dx(x)+f_3(x)fracdf_3dx(x))=0 iff f_1(x)fracdf_1dx(x)+f_2(x)fracdf_2dx(x)+f_3(x)fracdf_3dx(x)=0=f(x)cdotfracdfdx(x)Box$

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: $fracddx(f(x)+g(x))=fracddx f(x)+fracddx g(x)$ and $fracddx(af(x))=afracddxf(x)$ for functions $f$ and $g$ and a real number $a$.

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
$$2langle f(x),f'(x)rangle=0$$
The meaning of that is, that the position vector is perpendicular to the velocity vector.

Edit:
The second part is easy to obtain
$$fracddxlangle f(x),f(x)rangle=langle f'(x),f(x)rangle+langle f(x),f'(x)rangle=2langle f'(x),f(x)rangle=2langle f(x),f'(x)rangle$$
by product rule and symmetry.

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! 😉