Answer
$\dfrac{1}{2}$
Work Step by Step
We have: $ r=\sin 3\theta$ and $r (\theta=\dfrac{\pi}{4})=\dfrac{\sqrt 2}{2}$
$\dfrac{dr}{ d \theta}=3 \cos 3 \theta$ and $\dfrac{dr}{ d \theta}(\theta=\dfrac{\pi}{4})=3 \cos 3 (\dfrac{\pi}{4})=\dfrac{-3}{\sqrt 2}$
$\dfrac{dy}{dx}=\dfrac{dy/d \theta }{dx/d \theta }\\=\dfrac{r \cos \theta+\sin\theta \dfrac{dr}{ d \theta}}{- r \sin \theta+\cos\theta \dfrac{dr}{ d \theta}}$
Now,
$\dfrac{dy}{dx}=\dfrac{(\dfrac{\sqrt 2}{2}) (\dfrac{1}{\sqrt 2}) +\dfrac{1}{ \sqrt 2} \times (\dfrac{-3}{\sqrt 2})}{(\dfrac{-\sqrt 2}{2}) (\dfrac{1}{\sqrt 2}) +\dfrac{1}{ \sqrt 2} \times (\dfrac{-3}{\sqrt 2})}\\=\dfrac{1}{2}$
