Answer
$${y^2} + 6x - 9 = 0$$
Work Step by Step
$$\eqalign{
& r = \frac{3}{{1 + \cos \theta }} \cr
& {\text{Multiply both sides by }}\frac{1}{r} \cr
& 1 = \frac{3}{{r + r\cos \theta }} \cr
& r + r\cos \theta = 3 \cr
& {\text{Where }}r = \sqrt {{x^2} + {y^2}} {\text{ and }}r\cos \theta = x \cr
& \sqrt {{x^2} + {y^2}} + x = 3 \cr
& \sqrt {{x^2} + {y^2}} = 3 - x \cr
& {\text{Square both sides}} \cr
& {x^2} + {y^2} = 9 - 6x + {x^2} \cr
& {y^2} = 9 - 6x \cr
& {y^2} + 6x - 9 = 0 \cr} $$
