I had to plot the curve given by the parametric definition as follows. In attempting to simplify doing so, here is what I see holds between $x$ and $y$.
Plot the curve given by $$x=\frac{(t+2)^2}{t+1} ,y=\frac{(t-2)^2}{t-1}$$
$$\boxed{x=f(t)\\ y=-f(-t)}$$
Is there somehow we can use that to plot the graph easily. Any hints are appreciated. Thanks.
write $x$ as $\dfrac{(1+t+1)^2}{t+1} = t+1+2+\dfrac{1}{t+1}$ and $y = t-1 -2 + \dfrac{1}{t-1} $
$$\begin{align} x+y &= 2t \Big(\dfrac{t^2}{t^2-1} \Big) \\ y - x &= 6 - \dfrac{2}{t^2-1}\\ x+y &= t^3 \cdot (6+x-y) \end{align}$$
and now you can just put value of $t$ in terms of $x,y$
to get $$(8+x-y)^3 = (x+y)^2 (6+x-y) $$
here is what geogebra gave