$\displaystyle f(\vartheta)=\frac{3\sqrt{\vartheta^3}}{2\sin(2\va rtheta)}$

You will need to apply the chain rule and quotient rule here...

$\displaystyle f'(\vartheta)=\frac{\displaystyle \left[2\sin(2\vartheta)\cdot\frac{3}{2}(\vartheta^3)^{-\frac{1}{2}}\cdot3\vartheta^2\right]-\left[3\sqrt{\vartheta^3}\cdot 4\cos(2\vartheta)\right]}{4\sin^2(2\vartheta)}$$\displaystyle =\color{red}\boxed{\frac{9\vartheta^2\sin(2\varthe ta)-12\vartheta^3\cos(2\vartheta)}{4\sqrt{\vartheta^3} \sin^2(2\vartheta)}}$

I hope this makes sense!

--Chris