You have \(\displaystyle \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial v_\theta}{\partial r}\right)- \frac{V_\theta}{r^2}= 0\)

Now, by the product rule, \(\displaystyle \frac{\partial}{\partial r}\left(r\frac{\partial v}{\partial r}\right)= \frac{\partial r}{\partial r}\frac{\partial v_\theta}{\partial r}+ r \frac{\partial^2 v_\theta}{\partial r^2}\)\(\displaystyle = r\frac{\partial^2 v_\theta}{\partial r^2}+ \frac{\partial v_\theta}{\partial r}\).

\(\displaystyle \frac{1}{r}\) times that gives

\(\displaystyle \frac{\partial^2 v_\theta}{\partial r^2}+ \frac{1}{r}\frac{\partial v_\theta}{\partial r}\)

Subtracting \(\displaystyle \frac{v_\theta}{r^2}\) from that gives the second line:\(\displaystyle \frac{\partial^2 v_\theta}{\partial r^2}+ \frac{1}{r}\frac{\partial v_\theta}{\partial r}-\frac{v_\theta}{r^2}\)

To see that this is the same as the third line, use the quotient rule:

\(\displaystyle \frac{\partial}{\partial r}\frac{v_\theta}{r}= \frac{r\frac{\partial v_\theta}{\partial r}- v_\theta\frac{\partial r}{\partial r}}{r^2}\)\(\displaystyle = \frac{1}{r}\frac{\partial v_\theta}{\partial r}- \frac{v_\theta}{r^2}\)

and, of course, if we know that \(\displaystyle v_\theta\) is a function of r only, we can change the partial derivatives symbols to common derivatives: \(\displaystyle \partial\) to d.