Slope of The Tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$ Find slope of tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$
using $$\frac{dy}{dx} = \frac{\frac{dr}{dθ}\sin \theta + r \cos \theta}{\frac{dr}{dθ}\cos \theta-r \sin \theta}$$
I got
$$\frac {14\cos \theta \sin \theta}{7 \cos^2 \theta-7\sin \theta cos \theta}$$

When you have the equation in rectangular coordinates
$$
x^2+y^2=7y
$$
you can use implicit differentiation:
$$
2x+2yy'=7y'
$$
so
$$
y'=\frac{2x}{7-2y}
$$
For $\theta=\pi/6$, we have $r=7/2$, so $x=7\sqrt{3}/4$ and $y=7/4$; therefore the derivative at the point is
$$
y'(7\sqrt{3}/4)=\frac{2\dfrac{7\sqrt{3}}{4}}{7-2\dfrac{7}{4}}=\sqrt{3}
$$
If you want to do it directly with polar coordinates,
$$
dx=\cos\theta\,dr-r\sin\theta\,d\theta
\qquad
dy=\sin\theta\,dr+r\cos\theta\,d\theta
$$
which gives
$$
\frac{dy}{dx}
=\frac{\sin\theta\,dr+r\cos\theta\,d\theta}{\cos\theta\,dr-r\sin\theta\,d\theta}
=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}
=\tan2\theta
$$
Your derivation is almost correct: you have $\sin\theta\cos\theta$ at the bottom right, but it should be $7\sin^2\theta$.
