use polar coordiantes and the Poincare-Bendixson Theorem. I am searching for some help with this question in review for a upcoming exam, I have been having some trouble understanding the workings that have been provided and am just looking for some clarification.
The question says to show that,
$$x'=x+y-x(x^2+2y^2)$$
$$y'=-x+y-y(x^2+2y^2)$$
has at least one periodic solution. The hint given was to use polar coordiantes and the Poincare-Bendixson Theorem.
So using polar coordiates,
$$r'=\frac{xx'+yy'}{r}=-r(F(\theta)r^2-1)$$ 
So how did they calculate  $$-r(F(\theta)r^2-1)$$
and my final question is how does this next step come about? How did they calculate $F(\theta)$
$$F(\theta)=cos^4\theta+2sin^4\theta+3cos^2\theta sin^2\theta=1+sin^2\theta $$

We calculate
$$r'=\dfrac{xx'+yy'}{r} = \dfrac{1}{2} r^3 \cos 2 \theta-\dfrac{3}{2}r^3 + r = -r\left(\left(\dfrac{3}{2} - \dfrac{1}{2} \cos 2 \theta \right)r^2 - 1 \right) = -r(F(\theta)r^2-1)$$
where
$$F(\theta) = \left(\dfrac{3}{2} - \dfrac{1}{2} \cos 2 \theta\right) = 1 + \sin^2 \theta$$
Update
From $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta$, we have
$$\left(\dfrac{3}{2} - \dfrac{1}{2} \cos 2 \theta\right) = \dfrac{3}{2} -\dfrac{1}{2}\left(\cos^2 \theta - \sin^2 \theta\right) = \dfrac{3}{2} -\dfrac{1}{2}\left(1 - \sin^2 \theta\right) + \dfrac{1}{2} \sin^2 \theta = 1 + \sin^2 \theta$$
