$u(t) $ is differentiable vector function in $\mathbb{R}^3$ on $[a,b]$ and $u(t) \neq 0$ for all t.
$u^´(t)$ is the derivative of $u(t)$ and is orthogonal for $t \in (a,b)$ for all t
$\implies$ $u(t) \cdot u^´(t) = 0$
Also, its given that $u_0(t) = \frac{u(t)}{||u(t)||}$ is a radial unit vector.
To prove: $$u^´(t) = (u_0(t) \cdot u^´(t))u_0(t) + ||u(t)||u^´_0(t)$$
Thanks.
Since $u_0$ is in the direction of $u$, then $u_0\cdot u=0$, hence $(u_0\cdot u)u_0=0$, so the first term on the right hand side of your equation is zero. Next, note that $|u|=\sqrt{u\cdot u}$ so, $$\frac{d}{dt}|u|=\frac{d}{dt}\sqrt{u\cdot u}=\frac{u\cdot u'}{|u|^2}=0$$ from your argument above. Therefore, $$u_0'=\frac{|u|u'-u\frac{d}{dt}|u|}{|u|^2}=\frac{u'}{|u|}$$ so it follows that $|u|u_0'=u'$. Putting this together gives the desired result.