Answer
See graph and explanations.
Work Step by Step
Step 1. Convert the equation into polar coordinates using $r^2=x^2+y^2, x=r\cdot cos\theta, y=r\cdot sin\theta$, we have $r^2=(r^2-r^2cos^2\theta)^2$ which leads to $1=r^2(1-cos^2\theta)^2$ thus $r^2sin^4\theta=1$ or $r=\pm csc^2\theta$
Step 2. As the graph of $r=csc^2\theta$ is identical as $r=-csc^2\theta$, we graph the first one as shown in the figure.