Let $x (t) = t^2$ and $y (t) = t^3$, so that we can write $\phi (t) = (x (t), y (t))$. Note that $y (t) = \left( x(t) \right)^{3/2}$, which yields $\left (y (t)\right)^2 = \left( x (t)\right)^3$ or, equivalently $\left (y (t)\right)^2 - \left( x (t)\right)^3 = 0$. Hence, we have that $\phi (t)$ travels along the curve $C = \{ (x,y) \in \mathbb{R}^2 \mid y^2 - x^3 = 0\}$, which is depicted below
C.">
Note that for $t < 0$, $\phi (t)$ will lie on the 4th quadrant and approach the origin from infinity, whereas for $t > 0$ it will lie on the 1st quadrant and travel from the origin to infinity.
Every polynomial $f \in \mathbb{R}[X,Y]$ that vanishes on the curve $C$ can be factored as follows
$f (X, Y) = (Y^2 - X^3)^n \, g (X, Y)$
where $n \geq 1$ and $g \in \mathbb{R}[X,Y]$ does not vanish on $C$, and is thus divisible by $(Y^2 - X^3)$. I have no idea what property of a field $\mathbb{K}$ is required for the result to hold for the case $\phi : \mathbb{K} \to \mathbb{K}^2$.