Answer
$y=x^{3}-2x^{2},\ \ x\geq 0.$
Work Step by Step
From the parametric equation for x, we see that $x\geq 0.$
With $\left\{\begin{array}{l}
t^{6}=(t^{2})^{3},\\
t^{4}=(t^{2})^{2}
\end{array}\right.$
when we substitute into the parametric equation for y:
$\fbox{ $y=x^{3}-2x^{2},\ \ x\geq 0$}.$
$ y=x^{2}(x-2)\qquad\Rightarrow$ x-intercepts at x=0 and x=2.
To graph, create a table using several values for t, and then calculate $(x(t), y(t)),$ plotting the points as you go. Join with a smooth curve, noting the direction in which $t$ increases.
