Graphing $r^3\cos{3\theta}-1=0$ in the $xy$ plane.
How would I go about doing this? By computers I know that it looks like three hyperbola's. How could I see this without the help of a computer? Would the best way just be to plug in nice values for $\theta$ and solve for $r$ and get points until i can figure out what the graph looks like?
First, let us consider searching for when asymptotes occur, that is, when $r \rightarrow \infty$. Rearranging, we see this is equivalent to when $\frac{1}{\cos(3\theta)} \rightarrow \infty$.
This occurs at the roots of $\cos(3\theta)$, which in the domain $[0,2\pi]$ are given by $\theta_0 = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2},\frac{11\pi}{6}$.
Hence, we may sketch out our asymptotes at these angles, which in Cartesian form are given by $y = x\tan{\theta_0}$ for the above $\theta_0$ (e.g. $y = \frac{x}{\sqrt{3}}$)
Furthermore, we note that $r > 0$, so our curve can only exist in alternating regions between each asymptote.
Finally, we can consider searching for local minima of $r$, which occur for $3\theta = 2\pi n$, so we have minima at $\theta = 0,\frac{2\pi}{3}, \frac{4\pi}{3}$ where $r = 1$.
This gives you a set of key points, which together with the asymptotes, should enable you to recover the three 'quasi-hyperbolae' on pen and paper.
Let's rewrite
$$\cos(3 \theta) = \cos^3(\theta) - 3 \cos(\theta) \sin^2(\theta)$$
Recall that $x^2 + y^2 = r^2$, $x = r \cos(\theta)$, and $y = r\sin(\theta)$.
So, your original equation can be expressed in rectangular coordinates as:
$$(x^2 + y^2)^3 \frac{x^3 - 3xy^2}{(x^2 + y^2)^3} - 1 = 0 $$
Solving for $y$, we have:
$$y^2 = \frac{x^3 - 1}{3x} = \frac{x^2}{3} - \frac{1}{3x}$$
This last representation can let you see where the various asymptotes are, which may help you "see... that it looks like three hyperbolas." You can find it graphed here or illustrated in the screenshot of this graph below:
Prior knowledge of conics and polar curves would be useful to target this question, such as general shapes; e.g. rose curves, lemniscates etc. See Polar Curves On Brilliant.
The equation $r = \sec^{1/3}{3\theta}$ is an epispiral, since its inverse $r = \cos^{1/3}{3\theta}$ is equivalent of an odd rose curve. If we first analyse and draw the given rose curve above, it is not difficult to draw out the inverse.
Minima w.r.t to the radius
The rose curve $r = \cos^{1/3}{3\theta}$ has maximum radial points at $r = 1$, since $\max(\cos^{1/3}{n\theta}) = 1^{1/3} = 1$. Since, there are $n = 3$ petals, there are $3$ angles and they are evenly spread along the angular plane. $\cos({\theta = 0}) = 1$ being the first and a difference between each angle of $\dfrac{2\pi}{3}$, the other two would be $\theta = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$. Note that $r = 1$, hence the epispiral and the rose curve touch at these 3 angles, and this would its radial minima. We can figure out the $xy$ equivalent, using the parametric transformation $x = r\cos\theta$ and $y = r\sin\theta$. These would respectively be, $(x, y) = (1, 0), (-\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}), (-\dfrac{1}{2}, -\dfrac{\sqrt{3}}{2})$.
Asymptotes
Asymptotes will occur when the rose curve, $\cos{3\theta} = 0 \implies \dfrac{1}{\cos{3\theta}} \to \dfrac{1}{0} \to \infty$. Therefore, $\cos{3\theta} = 0$ over $[0, 2\pi]$ yields $6$ solutions; $\theta = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{3\pi}{2}, \dfrac{11\pi}{6}$. Since, $\theta$ represents the angle of made between the asymptote and the $x-$axis, we can figure out the cartesian equation of the asymptotes using $y = x \cdot \tan{\theta}$. These would respectively be $x = 0, y = \pm\dfrac{x}{\sqrt{3}}$.
This is all we require to draw the epispiral. It can be done as shown below:
Note the inclusion of the purple dotted circle. If one were to generalise the graph $r = a\sec^{1/3}{n\theta}$, notice that it always greater than the minimum circle bounding its rose inverse; $r^{-1} = a\cos^{1/3}{n\theta}$, since the two curves only meet at the roses' maximum radius. Hence, all we require then is the asymptotes and the minimum bounding circle $r = a$:
