$$x=\frac{3t}{1+t^3} , y=\frac{3t^2}{1+t^3} , t \neq -1,$$
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.
On
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 $$ x+y+1=0 $$
$$ x^3 + y^3 =\frac{27t^3}{(1+t^3)^3} +\frac{27t^6}{(1+t^3)^3} =\frac{27t^3(1+t^3)}{(1+t^3)^3}=\frac{27t^3}{(1+t^3)^2}$$ $$ x^3 + y^3 = 3xy$$