Showing that a curve $\Gamma^*$ is spherical I'm having troubles with this differential geometry problem:
Let $\Gamma : \overrightarrow{x} = \overrightarrow{x}(s)$ be a curve of $E^3$ with natural parameter $s$ which does not pass through the origin $O$. Now consider the curve 
$$\Gamma^* : \overrightarrow{x}^*(s) = \frac{\overrightarrow{x}(s)}{f(s)},$$
where $f(s) = \left|\overrightarrow{x}(s)\right|$. 
I need to show that $\Gamma^*$ is spherical and that $s$ is a natural parameter of $\Gamma^*$ if and only if $f^2 + (f')^2 = 1$. 

Usually when we want to show that a curve is spherical it is sufficient to show that 
$$\frac{\rho(s)}{\tau(s)} + (\rho'(s)\tau(s))' = 0$$
where $\rho(s) = \frac 1{\kappa(s)}$ is the radius of curvature and $\tau(s) = \frac 1{\sigma(s)}$ is the radius of torsion. However here I have no way of calculating these scalar quantities. I'm also lost with the second part of the question. 
 A: In general, a curve is spherical if and only if it lives on some sphere - in our case, the sphere of radius $1$ and center $0$. Since
$$\| \vec x ^* (s) \| = \left\| \frac {\vec x(s)} {f(s)} \right\| = \left\| \frac {\vec x(s)} {\| \vec x (s) \|} \right\| = \frac {\| \vec x(s) \|} {\| \vec x (s) \|} = 1 ,$$
it is clear that $\vec x ^* (s)$ sits on the unit sphere for every $s$. (Notice that the condition that $\Gamma$ should not pass through $0$ insures that $f(s) \ne 0$, so that we may divide by $f(s)$.)
Moving on to the second part, notice that if $\vec x ^* = \dfrac {\vec x} f$, then $\vec x = f \vec x^*$, so deriving with respect to $s$ gives
$$\vec x ' = f' \vec x ^* + f {\vec x ^*} ' ,$$
whence
$$\tag {*}\| \vec x ' \| ^2 = \langle \vec x ' , \vec x '\rangle = \langle f' \vec x ^* + f {\vec x ^*} ' , f' \vec x ^* + f {\vec x ^*} '\rangle = (f')^2 \langle \vec x ^*, \vec x ^* \rangle + 2 f' f \langle \vec x ^*, {\vec x ^*} '\rangle + f^2 \langle {\vec x ^*} ', {\vec x ^*} ' \rangle .$$
Since $\Gamma ^*$ is spherical, we have $\langle \vec x ^*, \vec x ^* \rangle = 1$. But we can squeeze even more out of this fact! Deriving it with respect to $s$ gives
$$0 = \langle {\vec x ^*} ' , {\vec x ^*} \rangle + \langle {\vec x ^*} , {\vec x ^*} ' \rangle = 2 \langle {\vec x ^*} ' , {\vec x ^*} \rangle ,$$
whence $\langle {\vec x ^*} ' , {\vec x ^*} \rangle = 0$, a result that we shall also plug in $(*)$.
Finally, remember that $\Gamma$ carries the natural parametrization, which implies that $\| \vec x ' \| = 1$, yet another fact that we shall plug in $(*)$.
After all these remarks, then, formula $(*)$ can be rewritten as
$$\tag {**} 1 = (f')^2 + f^2 \langle {\vec x ^*} ', {\vec x ^*} ' \rangle .$$
Things are easy now. If $\Gamma ^*$ carries the natural parametrization, then $\| {\vec x ^*} ' \| = 1$, so $(**)$ becomes $1 = (f')^2 + f^2$. Conversely, if $1 = (f')^2 + f^2$, then $(**)$ becomes $f^2 = f^2 \langle {\vec x ^*} ', {\vec x ^*} ' \rangle$ and, since $\Gamma$ does not pass through $0$, we have $f(s) \ne 0 \ \forall s$ so we may divide by $f^2$ and obtain $1 = \langle {\vec x ^*} ', {\vec x ^*} ' \rangle = \| {\vec x ^*} ' \| ^2$, whence $\| {\vec x ^*} ' \| = 1$, showing that $\Gamma ^*$ carries the natural parametrization.

It is interesting to notice that if you had started your proof by deriving $\vec x ^* = \dfrac {\vec x} f$, you would have obtained 
$$\| {\vec x ^*} ' \| ^2 = \frac {(f') ^2} {f^4} \| x \| ^2 - 2 \frac {f'} {f^3} \langle \vec x, \vec x ' \rangle + \frac 1 {f^2} \| \vec x ' \| ^2$$
and the only thing that you could have done here would have been replacing $\| \vec x ' \|$ by $1$ (because of $\Gamma$ being naturally parametrized), and then you would have been stuck.
The thing that you learn from here is that the most straightforward and intuitive approach to a problem might quickly get you stuck, while a more convoluted approach might lead to the solution. (Unfortunately, this happens almost always in mathematics, which is why work in this field is so difficult.)
