Answer
$-27x^2+9y^2-81+108x=0$
Work Step by Step
After multiplying with the denominator of the fraction: $r(3-6\cos{\theta})=9\\3r-6r\cos{\theta}=9\\3r=9-6r\cos{\theta}\\9r^2=(9-6r\cos{\theta})^2$.
We know that $r^2=x^2+y^2$ and that $x=r\cos\theta,y=r\sin\theta$.
Hence $9r^2=(9-6r\cos{\theta})^2$ becomes: $9(x^2+y^2)=(9-6x)^2\\9x^2+9y^2=81-108x+36x^2\\-27x^2+9y^2-81+108x=0$
