= Solution
Take $a>0$ for the sketch; changing its sign rotates the picture by $\pi$. The <parametric curve> passes through $(0,0)$ at $t=0$ and through $(3a/2,3a/2)$ at $t=1$. As $t\to\pm\infty$ it returns towards the origin, since $x\sim3a/t^2$ and $y\sim3a/t$. The positive-$t$ branch makes a loop in the first quadrant, leaving the origin tangent to the positive $x$ axis and returning tangent to the positive $y$ axis.
For a more precise sketch, differentiation gives
$$
x'(t)=\frac{3a(1-2t^3)}{(1+t^3)^2},\qquad
y'(t)=\frac{3at(2-t^3)}{(1+t^3)^2}.
$$
The loop has its largest $x$ at $t=2^{-1/3}$ and largest $y$ at $t=2^{1/3}$. Their coordinates are $a(2^{2/3},2^{1/3})$ and its reflection in $y=x$.
The parameter value $t=-1$ is excluded. Cancelling the common factor in the sum gives
$$
x+y=\frac{3at}{t^2-t+1}\longrightarrow-a,
\qquad x-y\sim-\frac{2a}{t+1}.
$$
Thus the <asymptote> is $\boxed{x+y+a=0}$. As $t\to-1^-$, $x\to+\infty$ and $y\to-\infty$; as $t\to-1^+$, $x\to-\infty$ and $y\to+\infty$. The branch $t<-1$ approaches the origin from the fourth quadrant as $t\to-\infty$, while $-1<t<0$ connects the other infinite branch to the origin through the second quadrant.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2-folium.png]
{title=Folium of Descartes with parameter directions, marked points, and its oblique asymptote}
This is the <folium of Descartes>. If $a=0$, the parametrization degenerates to the origin for every defined parameter value and there is no nontrivial loop.
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