Chain rule differentiation 
Given that $\overline{r} = \frac{r}{r_0}$ and $u = \frac{1}{\overline{r}}$, show that 
  $$r \frac{d^2r}{d\theta^2} - 2\left(\frac{dr}{d\theta} \right)^2 = r^2 - 3\frac{Gm}{c^2}r$$
  is equivalent to the following;
  $$\frac{d^2u}{d\theta^2} + u = \epsilon u^2 : \epsilon = \frac{3Gm}{r_0c^2}
$$ 

I understand that we can start with the working $\frac{dr}{d\theta} = r_0 \frac{d({\frac{1}{u})}}{d\theta}$ but then I cannot see how you would move further on. 
Any help would be appreciated. 

\begin{equation}
r=\frac{r_{0}}{u}
\end{equation}
\begin{equation}
\frac{dr}{d\theta}=\frac{d}{d\theta}(\frac{r_{0}}{u}) 
\end{equation}
Applying chain rule of differentiation you'll get
\begin{equation}
\frac{dr}{d\theta}=\frac{-r_{0}}{u^{2}}\frac{du}{d\theta}
\end{equation}
\begin{equation}
\frac{d^{2}r}{d\theta^{2}}=\frac{d}{d\theta}(\frac{-r_{0}}{u^{2}}\frac{du}{d\theta})
\end{equation}
Here you'll have to use product rule of differentiation that is, if independent variable of differentiation is $t$ then,
\begin{equation}
\frac{d}{dt}(uv)=u\frac{dv}{dt}+v\frac{du}{dt}
\end{equation}
\begin{equation}
\therefore \frac{d^{2}r}{d\theta^{2}}=\frac{2r_{0}}{u^{3}}\left(\frac{du}{d\theta}\right)^{2}-\frac{r_{0}}{u^{2}}\frac{d}{d\theta}\left(\frac{du}{d\theta}\right)
\end{equation}
\begin{equation}
\therefore \frac{d^{2}r}{d\theta^{2}}=\frac{2r_{0}}{u^{3}}\left(\frac{du}{d\theta}\right)^{2}-\frac{r_{0}}{u^{2}}\left(\frac{d^{2}u}{d\theta^{2}}\right)
\end{equation}
Multiplying on both sides by r in above equation we get,
\begin{equation}
r\frac{d^{2}r}{d\theta^{2}}=\frac{2r_{0}r}{u^{3}}\left(\frac{du}{d\theta}\right)^{2}-\frac{r_{0}r}{u^{2}}\left(\frac{d^{2}u}{d\theta^{2}}\right)
\end{equation}
Substitue $r$ in terms of $u$
\begin{equation}
r\frac{d^{2}r}{d\theta^{2}}=\frac{2r_{0}^{2}}{u^{4}}\left(\frac{du}{d\theta}\right)^{2}-\frac{r_{0}^{2}}{u^{3}}\left(\frac{d^{2}u}{d\theta^{2}}\right)
\end{equation}
And 
\begin{equation}
\left(\frac{dr}{d\theta}\right)^2=\frac{r_{0}^{2}}{u^{4}}\left(\frac{du}{d\theta}\right)^2
\end{equation}
Thus,
\begin{equation}
2\left(\frac{dr}{d\theta}\right)^2=\frac{2r_{0}^{2}}{u^{4}}\left(\frac{du}{d\theta}\right)^2
\end{equation}
Subtract appropriate equations and use original differential equation in terms of $r$
\begin{equation}
r\frac{d^{2}r}{d\theta^{2}}-2\left(\frac{dr}{d\theta}\right)^2=-\frac{r_{0}^{2}}{u^{3}}\left(\frac{d^{2}u}{d\theta^{2}}\right)
\end{equation}
\begin{equation}
r^{2}-\frac{3GMr}{c^{2}}=-\frac{r_{0}^{2}}{u^{3}}\left(\frac{d^{2}u}{d\theta^{2}}\right)
\end{equation}
Again on Left hand side convert $r$ to $u$
\begin{equation}
\frac{r_0^{2}}{u^{2}}-\frac{3GMr_0}{c^{2}u}=-\frac{r_{0}^{2}}{u^{3}}\left(\frac{d^{2}u}{d\theta^{2}}\right)
\end{equation}
Divide by $-r_0^{2}$ and multiply by $u^3$ on both sides
\begin{equation}
-u+\frac{3GMu^2}{c^{2}r_0}=\frac{d^{2}u}{d\theta^{2}}
\end{equation}
Rearranging and putting value of $\epsilon$ you'll have
\begin{equation}
\frac{d^{2}u}{d\theta^{2}}+u=\epsilon u^{2}
\end{equation}
(Note: This kind of manipulation is useful particularly when you're trying to find orbit equation in classical mechanics problem, I'm sure you're doing that only)
