[tex]\mathbf r(t)=x(t)\,\mathbf i+y(t)\,\mathbf j+z(t)\,\mathbf k=t\,\mathbf i+(4t-t^2)\,\mathbf k[/tex]
[tex]\implies \begin{cases}x(t)=t\\y(t)=0\\z(t)=4t-t^2\end{cases}[/tex]
So the paraboloid is given by
[tex]z(t)=x(t)^2y(t)^2\iff 4t-t^2=0[/tex]
which holds for [tex]t^2-4t=t(t-4)=0[/tex], i.e. when [tex]t=0[/tex] and [tex]t=4[/tex].
To make sure this is correct, plug in [tex]t=0[/tex] and [tex]t=4[/tex] to find out the value of [tex]\mathbf r(t)[/tex], and see if these points are on the paraboloid.
[tex]\mathbf r(0)=0\,\mathbf i+0\,\mathbf j+0\,\mathbf k\implies (0,0,0)[/tex]
[tex]0=0^2\times0^2[/tex]
[tex]\mathbf r(4)=4\,\mathbf i+0\,\mathbf j+(4\times4-4^2)\,\mathbf k\implies (4,0,0)[/tex]
[tex]0=4^2\times0^2[/tex]
Both of these hold, so the answer is correct.
