Answer
$\left\{\begin{array}{ll}
x=3\sin t, & \\
y=9\sin^{2}t, &
\end{array}\right.\quad t\in[0,+\infty)$
Work Step by Step
Using the fact that $\sin t\in[-1,1]$
we can keep x between $-3$ and $3$ with $x=3\sin t$.
We want x to start at 0, so we start with t=0 (restrict t to $[0,+\infty)$ )
Substituting x in $y=x^{2}$,
$y=(3\sin t)^{2}$
$y=9\sin^{2}t$
So, we can have the following:
$\left\{\begin{array}{ll}
x=3\sin t, & \\
y=9\sin^{2}t, &
\end{array}\right.\quad t\in[0,+\infty)$
