I posted this question all the way back in 2014...and yes, I'm still trying to solve it.
To refresh some memories, here is the little demon in all its glory...
$\displaystyle x \cos \theta + y \sin \theta = \cos (3 \theta)$; $x \sin \theta - y \cos \theta = 3 \sin (3 \theta)$
I actually have some progress...I finally reached some progress when using $$x^2 + y^2 = \cos^2 3 \theta + 9 \sin^2 3 \theta$$ which boils down neatly to $$ 5-(x^2+y^2) = 4 \cos 6 \theta$$ - I then multiplied the first equation by 2 and multiplied that by the second to get
$$(x^2-y^2)(2 \sin \theta \cos \theta) + 2xy (\cos^2 \theta - \sin^2 \theta) = 6 \sin 6 \theta.$$ I would think subsituting $x = \cos \theta$ or $y = \sin \theta$ might give some insight, but I'm not sure.
Is this close? Am I missing something?
(The answer for this is still $\displaystyle(x^2 + y^2)(x^2+y^2+18)+8x(x^2-3y^2) = 27$.)