= Solution
Follow a fluid particle released at time $s$, and use $\tau$ for its running time. Its <pathline> satisfies $dx/d\tau=1$, $dy/d\tau=\tau x$ and $dz/d\tau=0$, with zero position at $\tau=s$. Hence $x(\tau)=\tau-s$, $z(\tau)=0$, and
$$
y(t)=\int_s^t\tau(\tau-s)d\tau=\frac{t^3}{3}-\frac{st^2}{2}+\frac{s^3}{6}.
$$
At observation time $t$, set $x=t-s$. Substitution gives the <streakline>
$$
\boxed{(x,y,z)=\left(t-s,\ \frac12t(t-s)^2-\frac16(t-s)^3,\ 0\right),\quad y=\frac12tx^2-\frac16x^3.}
$$
The curve is parameterized by release times $s\leq t$. If release starts at time zero, only $0\leq x\leq t$ is occupied. This is a <streakline>, the positions of different particles released at one fixed point, rather than an instantaneous <streamline> of this unsteady flow.
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