If $\displaystyle r$ and $\displaystyle \theta$ are funtions of $\displaystyle t$, does $\displaystyle \frac{d^2 r}{dt^2}\bigg / \left ( \frac{d\theta}{dt}\right ) ^2$ equal $\displaystyle \frac{d^2 r}{d\theta ^2}$?
If $\displaystyle r$ and $\displaystyle \theta$ are funtions of $\displaystyle t$, does $\displaystyle \frac{d^2 r}{dt^2}\bigg / \left ( \frac{d\theta}{dt}\right ) ^2$ equal $\displaystyle \frac{d^2 r}{d\theta ^2}$?
No. While the first derivative can be "treated like" a fraction, the second derivative cannot.
For example, take $\displaystyle r= t^2$ and $\displaystyle \theta= t^4$. Then $\displaystyle \frac{dr}{dt}= 2t$, $\displaystyle \frac{d\theta}{dt}= 4t^3$, $\displaystyle \frac{d^2r}{dt^2}= 2$ and $\displaystyle \frac{d^2\theta}{dt^2}= 12t^2$.
Now it is easy to see that $\displaystyle \theta= (t^2)^2= r^2$ so $\displaystyle \frac{d\theta}{dr}= 2r= 2t^2= \frac{4t^3}{2t}$$\displaystyle = \frac{\frac{d\theta}{dt}}{\frac{dr}{dt}}$.
But $\displaystyle \frac{d^2\theta}{dr^2}= 2$ while $\displaystyle \frac{\frac{d^2\theta}{dt^2}}{\frac{d^2r}{dt^2}}= \frac{12t^2}{2}= 6t^2$.