An asymptote is going to mean that $r\rightarrow\infty$, which happens when the denominator approaches zero. Vertical or horizontal means that $\theta$ is a multiple of $\pi/2$ (vertical for even, horizontal for odd multiples). The only asymptote would therefore be when $\theta\rightarrow\pi$, which would be a horizontal one above (and parallel to) the negative $x$ axis.
Now as $\theta\rightarrow\pi$,
$$
y = r\sin\theta
= \frac{\theta\sin\theta}{\pi-\theta}
= \frac{\theta\sin(\pi-\theta)}{\pi-\theta}
\rightarrow \theta
\rightarrow \pi
$$
so the horizontal asymptote is $y=\pi$.
For visualization, we can reason that as $\theta$ increases from $0$ to $\pi$, $r$ increases from $0$ to $\infty$, crossing $1$ at the positive $y$ axis. Here are some plots for $\theta\in[0,.6\pi]$ and $[0,.99\pi]$ made with sage (online):
x_coords = [t/(pi-t)*cos(t) for t in srange(0, 0.6*pi, 0.02)]
y_coords = [t/(pi-t)*sin(t) for t in srange(0, 0.6*pi, 0.02)]
list_plot(zip(x_coords, y_coords))
x_coords = [t/(pi-t)*cos(t) for t in srange(0, 0.99*pi, 0.02)]
y_coords = [t/(pi-t)*sin(t) for t in srange(0, 0.99*pi, 0.02)]
list_plot(zip(x_coords, y_coords))
Horizontal and vertical tangents can be found by finding the roots of
$\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ respectively (assuming
that they are never simultaneously zero, in which case we'd have to
resort to higher order derivatives). A simple way to find these roots
is to observe that
$$r = \left(\frac{\pi}{\theta}-1\right)^{-1}
\implies \frac{dr}{d\theta}
= -\left(\frac{\pi}{\theta}-1\right)^{-2} \cdot
\left(-\pi\theta^{-2}\right)
= \frac{\pi}{(\theta-\pi)^2}
$$
so that
$$
0 = \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta
\iff
\tan\theta = \frac{1}{r} \frac{dr}{d\theta} = \frac{\pi}{\theta(\pi-\theta)}
$$
and
$$
0 = \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta
\iff
-\cot\theta = \frac{1}{r} \frac{dr}{d\theta} = \frac{\pi}{\theta(\pi-\theta)}
$$
which show us that they never vanish simultaneously.
From the shape of $\frac{\pi}{\theta(\pi-\theta)}$,
which is symmetric and positive on $(0,\pi)$
with global minimum at $\frac{\pi}{2}$
and vertical asymptotes at the endpoints,
we see that it will meet $\tan\theta$
at exactly one point $\theta_0\in(0,\frac{\pi}{2})$;
thus, there is one unique vertical tangent to our curve.
From the graph above, we might estimate
$$
\tan\theta_0\approx\frac{.4}{.25}=1.6
\implies
\theta_0
\approx \frac{\pi}{2}-\sqrt{\frac{\pi}{2}\left(\frac{\pi}{2}-\frac{5}{4}\right)}
\approx .8609
$$
where we used the quadratic equation
to deduce $\theta$ from $\tan\theta$.
However, the actual solution seems to be
closer to $0.97803904765198235$:
t=var('t'); find_root(t*(pi-t)*tan(t)==pi, 0, pi/2)
For horizontal tangents, the same function
$\frac{\pi}{\theta(\pi-\theta)}$ must now meet
$-\cot\theta$, which is equivalent to the equation
$$\tan\theta=\theta\left(1-\frac{\pi}{\theta}\right) $$
which on $[0,\pi]$ has only solutions at the two endpoints,
i.e., on our curve, at the origin and
at the horizontal asymptote found above.
$$\dfrac{dy}{d\theta}\cdot \frac{d\theta}{dx} = \dfrac{dy}{dx}$$ and we want to know when $\dfrac{dy}{dx} = 0$.
$$\frac{dx}{d\theta} = \frac{\cos3\theta}{r},\quad \frac{dy}{d\theta} = \frac{\sin 3\theta}{r} \implies \dfrac{dy}{dx} = \frac{\sin{3\theta}}{r}\cdot \dfrac{r}{\cos 3\theta}$$
Now solve for when $$\require{cancel} \dfrac{dy}{dx} = \frac{\sin{3\theta}}{\cancel{r}}\cdot \dfrac{\cancel{r}}{\cos 3\theta} = = \tan(3\theta)= 0$$
That's going to happen when $\sin 3\theta = 0$ (and note that when $\sin{3\theta} = 0 \iff 3\theta = k \pi \iff \theta = k\pi/3\;$ ($k$ any integer), we have that $\cos 3\theta = \pm 1\iff 3\theta = k\pi \iff \theta = k\pi/3$.)
The points at which the tangents are horizontal and/or vertical are given in polar coordinates, $(r, \theta)$.
Best Answer
Dividing by $\cos(\theta)$: Note the $\cos(\theta)$ could be zero, in which case it doesn't make sense to straight away divide $\cos(\theta)$.
There so there should be some case analysis:
Then $$2\sin(\theta) = 1 \\ \theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$$
Then $$\theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$$ However, by substituting into the equation for the derivative, the denominator, $\sin^2\theta−\cos^2\theta−\sin\theta$ becomes $0$ when $\theta = \dfrac{\pi}{2}$. Hence, the derivative does not exist at $\theta = \dfrac{\pi}{2}$.
Therefore we dismiss the root $\dfrac{\pi}{2}$, so the roots to the derivative, and hence the horizontal tangents to the curve are: $$\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$$.