Prove that $y^2+x^2+c^2x^2(1-x^2)=1$ is closed as a path.

Question

How can I prove that the curve defined implicitly by $y^2+x^2+c^2x^2(1-x^2)=1$ and passes through the point (0,1) is closed as path when $c\in(0,1)$. Where closed means that for a parametrization $f(t):[0,1]\to \mathbf{R}^2$ of the curve, we have $f(0)=f(1)$.