Parameterization of a Strophoid: $y^2=x^3+x^2$ 
  
*The curve $C$ is given by the Cartesian equation: $y^2=x^3+x^2$.
  Show that the function $\overrightarrow{r}:\mathbb{R^2} \rightarrow \mathbb{R}$ defined by $\overrightarrow{r}(t)=(t^2-1,t^3-t)$, represents the parameterization of the curve $C$.
  

I solved this question (and correct me if I'm wrong) by the following:
let $x=t^2-1$, then $y^2=(t^2-1)^3-(t^2-1)^2$
$y^2=(t^6-3t^3+3t^2-1)+(t^3-2t^2+1)$
$y^2=t^6-2t^4+t^2=t^2(t^4-2t^2+1)=t^2(t^2-1)^2$
then we get:
$y=t(t^2-1)=t^3-t$
My question is, how do I find the parameterization of $C$ if the question hadn't already defined $\overrightarrow{r}(t)=(t^2-1,t^3-t)$?
Knowing that $x=t^2-1$ made the question trivial, is there a way to figure what $x$ is equal to?
 A: As Alex Francisco commented, usually the plane curve defined by a cubic polynomial equation is an elliptic curve. Those cannot be parametrized by rational functions. The exceptions are the cubic curves with a singular point. Here you see that both partial derivatives vanish at the origin, a point on your curve, so we have a singularity at the origin. 
The presence of a singularity changes everything (in the language of algebraic geometry the genus of the curve changes from one to zero). A recurring trick in algebraic geometry is to "blow up the singularity". We use the ratio $t=y/x$ as a parameter $t$ (if the singularity were at the point $(x_0,y_0)$ instead of the origin we would use $t=(y-y_0)/(x-x_0)$ instead. The idea is that using $t$ instead of $y$ simplifies the equation on such occasions. 
Let's see. If $t=y/x$ then $y=tx$. Plugging this into your equation gives
$$
t^2x^2=y^2=x^3+x^2.
$$
Due to the singularity only terms quadratic (or higher) in $x$ and $y$ appeared in the original equation. Therefore this substitution made both sides divisible by $x^2$. If we assume (temporarily) that $x\neq0$ we can cancel that factor $x^2$ and end up with
$$
t^2=x+1.
$$
Lo and behold! We just solved that $x=t^2-1$! Consequently
$$
y=tx=t(t^2-1).
$$
Observe that if $x=0$ then $y=0$ is the only possibility. We do get $x=0$ from two different choices of $t$, namely $t=\pm1$. Both of those values, of course, give us the origin as the point $(x(t),y(t))$. If you plotted this curve you will see that it has two branches throught the origin. When $t\approx1$ you get points on the ascending branch (with slope close to $+1$) but with $t\approx-1$ you get points on the descending branch. Given that $t$ is the slope of the line segment from the origin to the point $(x,y)$ this was to be expected. Blowing up separates those two branches - in the $(x,t)$ coordinate system the origin is no longer "a double point", and the singularity is gone.

As an exercise I invite you to use the same trick and derive a rational parametrization for the points of the Folium of Descartes
$$
x^3+y^3=3xy.
$$
It also has a singularity at the origin.
