Eliminate the parameter from the parametric equations $$x=\frac{3t}{1+t^3} ,  y=\frac{3t^2}{1+t^3}  , t \neq -1,$$
and hence find an ordinary equation in x and y for this curve, The parameter t can be interpreted as the slope of the line joining the general point $(x,y)$ to the origin. Sketch the curve and show that the line $x+y=-1$ is an asymptote.

Note that
$$
t=\frac yx
$$
Then
$$
x=\frac{3\frac yx}{1+\frac{y^3}{x^3}}=\frac{3x^2y}{x^3+y^3}\implies x^3+y^3-3xy=0
$$
As $x$ gets big, note that the major terms are $x^3$ and $y^3$, so if we scale $x$ and $y$ by $\lambda$, we get
$$
x^3+y^3-\frac3\lambda xy=0\to0=x^3+y^3=(x+y)(x^2-xy+y^2)
$$
Since there is no solution to $x^2-xy+y^2=0$ we are left with the scaled asymptote
$$
x+y=0
$$
If we scale back, and check with the original formulae, we get
$$
x+y\to\lim_{t\to-1}\frac{3t+3t^2}{1+t^3}=\lim_{t\to-1}\frac{3+6t}{3t^2}=-1
$$
Thus, we get the unscaled asymptote
$$
x+y+1=0
$$
