Why is $\frac{d}{dt}(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}) \cdot \vec{u} =\frac{m \vec u}{(1-\frac{u^2}{c^2})^{3/2}} \cdot \frac{d \vec u}{dt}$? Given that $\vec u(t)$, $u=|\vec u(t) |$ and $c$, $m$ are constants, how does one get from the LHS of the following equation to the RHS? 

$$\frac{d}{dt}\left(\frac{m \vec{u}}{\sqrt{1-\frac{u^2}{c^2}}}\right) \cdot \vec{u} =\frac{m \vec u}{\left(1-\frac{u^2}{c^2}\right)^{3/2}} \cdot \frac{d \vec u}{dt}$$

What is the differentiation rule used?
It's just the usual product and chain rules, but it helps to work with indices and sum implicitly when they repeat. I'll set $c=1$ to simplify notation:$$\frac{d}{dt}\left((1-u_ju_j)^{-1/2}mu_i\right)u_i=\left(u_k\dot{u}_k(1-u_ju_j)^{-3/2}mu_i+(1-u_ju_j)^{-1/2}m\dot{u}_i\right)u_i\\=(1-u_ju_j)^{-3/2}mu_i\dot{u}_i\left(u_iu_i+1-u_iu_i\right)=(1-u_ju_j)^{-3/2}mu_i\dot{u}_i.$$
The usual one. Try this: $u^2=(u,u)$ and then $\dfrac{d}{dt}u^2= 2(u',u).$