Where does this formula come from? I am doing revision for my module stellar & galactic astrophysics and have come upon this formula which I cannot seem to derive. Could someone please explain where it comes from?
"For an axisymmetric system, $\varphi$ component of equation of motion is:
$2 \dot{r} \dot{\varphi} + r \ddot{\varphi} = 0$"
I have the formula below in my textbook, so I assume it comes from there, but don't see how.

 A: In an axisymmetric system, the gravitational force on a particle is going to be radial. It therefore exerts zero torque about the origin, so angular momentum about the origin is conserved:
$$L = mr^2\dot{\varphi} = \text{const.} $$
Now differentiate both sides with respect to time to get
$$ 2mr\dot{r}\dot{\varphi} + mr^2\ddot{\varphi} = 0,$$
which upon dividing by $mr$ is is the result you want.
Alternatively, we can get the equation of motion from the Lagrangian for a central force
$$ \mathcal{L} = \tfrac{1}{2}m\dot{r}^2 + \tfrac{1}{2}mr^2\dot{\varphi}^2 - U(r).$$
The Euler-Lagrange equation for the $\varphi$ coordinate is
\begin{align}
\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{\varphi}} &= \frac{\partial \mathcal{L}}{\partial \varphi}\\
\implies \frac{d}{dt} (mr^2\dot{\varphi}) &= 0
\end{align}
which gives the same equations as before.

For the radial equation of motion, we can use the same ideas. If you've seen the Lagrangian approach before, it's easy ― just write down the Euler-Lagrange equation for the radial coordinate. If not, you can use the other conserved quantity that we know about, which is energy. We have
$$E = \tfrac{1}{2}m\dot{r}^2 + \tfrac{1}{2}mr^2\dot{\varphi}^2 + m\Phi(r)$$
where $\Phi$ is the gravitational potential (which needs to be multiplied by $m$ to get the potential energy). Now differentiate with respect to time:
$$ m\dot{r}\ddot{r} + mr\dot{r}\dot{\varphi}^2 + mr^2\dot{\varphi}\ddot{\varphi} + m\dot{r}\frac{d\Phi}{dr} = 0$$
and eliminate $\ddot{\varphi}$ to get the radial equation.
This link contains some MIT notes detailing another method, and the equations of motion you need are at the top of page 6.
A: $$\dot{\vec{r}} = \dot{r}\hat{e_r}+{r}\dot{\phi}\hat{e_{\phi}}$$
$$\ddot{\vec{r}} = (\ddot{r}-r\dot{\phi}^2)\hat{e_r}+({2r}\dot{\phi}+r\ddot{\phi})\hat{e_{\phi}}$$
$$m\ddot{\vec{r}} = m(\ddot{r}-r\dot{\phi}^2)\hat{e_r}+m({2r}\dot{\phi}+r\ddot{\phi})\hat{e_{\phi}}=-G\frac{Mm}{r^2}\hat{e_r}$$
thus
$$\ddot{\vec{r}} = -\nabla \Phi$$
