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.