Answer
$$\dfrac{4}{3}$$
Work Step by Step
We have: $ r=4-3 \sin \theta$ and $r (\theta=\pi)=4- 3 \sin \pi=4$
$\dfrac{dr}{ d \theta}=-3 \cos \theta$ and $\dfrac{dr}{ d \theta}(\theta=\pi)=-3 \cos \pi=3$
$\dfrac{dy}{dx}=\dfrac{dy/d \theta }{dx/d \theta }\\=\dfrac{r \cos \theta+\sin\theta \dfrac{dr}{ d \theta}}{- r \sin \theta+\cos\theta \dfrac{dr}{ d \theta}}$
Now,
$\dfrac{dy}{dx}=\dfrac{(4)(-1)+(0)(3)}{-4(0)+(-1)(3)}\\=\dfrac{-4}{-3}\\=\dfrac{4}{3}$
