<s>[INST] Polar to cartesian form of r=sin(3θ) $$\sin(2θ) = \sinθ\cosθ + \cosθ\sinθ = 2\sinθ\cosθ $$
$$\cos(2θ) = \cosθ\cosθ - \sinθ\sinθ = \cos²θ - \sin²θ $$
$$\sin(3θ) = \sin(2θ+θ) $$
$$= \sin(2θ)\cosθ + \cos(2θ)\sinθ $$
$$= 2\sinθ\cos²θ + \cos²θ\sinθ - \sin³θ 
$$ $$= 2\sinθ(1-\sin²θ) + (1-\sin²θ)\sinθ - \sin³θ 
$$ $$= 2\sinθ - 2\sin³θ + \sinθ - \sin³θ - \sin³θ 
$$ $$r=\sin(3θ)= 3\sinθ - 4\sin³θ$$
How to deal with the differing exponents? I could add an $r$ to both sides or $r^3$ but not both. 
 $$r^4= 3r^3\sinθ - 4r^3\sin^{3}θ   $$
 $$r^4= 3r^2y - 4y^3$$
 $$(x^2+y^2)^{4/2}= 3(x^2+y^2)^{2/2}y - 4y^3$$
 $$(x^2+y^2)^{2}= 3(x^2+y^2)y - 4y^3$$
 $$(x^2+y^2)^{2}= 3yx^2-y^3$$
 [/INST] Solution: 
$$r=3\sin(\theta)-4\sin^3(\theta)=\sin(\theta)\{ 4\cos^2(\theta)-1\}$$
Now put: $\cos(\theta)=x/r$ , $\sin(\theta)=y/r$ and $r=(x^2+y^2)^{1/2}$.
This will give you: $$r=\frac{y(4x^2-r^2)}{r^3}$$ which implies $$(x^2+y^2)^2=4x^2y-y(x^2+y^2)=3x^2y-y^3$$
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