Let's set $c=1$ for simplicity.
Using your observations, it suffices to show that (just combine the second and third equations you write down) $$ \dot \gamma = \vec u \cdot \frac{d}{dt}(\gamma \vec u) $$$$ \dot \gamma = \vec u \cdot \frac{d}{dt}(\gamma \vec u). $$ To prove this, the following facts are useful: $$ \dot \gamma = \gamma^3\vec u \cdot\dot{\vec u}, \qquad \gamma^2\vec u^2 +1 = \gamma^2 $$$$ \dot \gamma = \gamma^3\vec u \cdot\dot{\vec u}, \qquad \gamma^2\vec u^2 +1 = \gamma^2. $$ Now just compute \begin{align} \vec u \cdot \frac{d}{dt}(\gamma \vec u) &=\vec u \cdot (\dot \gamma \vec u + \gamma \dot{\vec u}) \\ &=\vec u \cdot (\gamma^3(\vec u \cdot \dot{\vec u})\vec u + \gamma \dot{\vec u}) \\ &= \gamma \vec u \cdot \dot{\vec u}(\gamma^2 \vec u^2 + 1) \\ &= \gamma^3\vec u \cdot\dot {\vec u} \\ &= \dot \gamma \end{align}