Answer
$$
x=3 t- t^{3}, \quad y=3 t^{2}
$$
the length of the loop of the curve is
$$
\begin{aligned} L &=\int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t
\\
&=12 \sqrt{3} \end{aligned}
$$
Work Step by Step
$$
x=3 t- t^{3}, \quad y=3 t^{2}
$$
then
$$
d x / d t=3 -3 t ^{2} , \quad d y / d t=6 t
$$
so
$$
\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2} = (3 -3 t ^{2} )^{2}+ (6t)^{2} =(3+3t^{2})^{2 }
$$
Thus the length of the loop of the curve is
$$
\begin{aligned} L &=\int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t
\\
&=\int_{-\sqrt{3}}^{\sqrt{3}}\left(3+3 t^{2}\right) d t \\
&=2 \int_{0}^{\sqrt{3}}\left(3+3 t^{2}\right) d t \\
&=2\left[3 t+t^{3}\right]_{0}^{\sqrt{3}} \\ &=2(3 \sqrt{3}+3 \sqrt{3}) \\
&=12 \sqrt{3} \end{aligned}
$$
