Eliminate $t$ from the equations: $x = \frac{1}{t} - t \, , \, y = \frac{1}{t^2} - 1$ $$x = \frac{1}{t} - t \tag{1}$$
$$y = \frac{1}{t^2} - 1 \tag{2}$$
$$\frac{1}{t^2} = y + 1$$
$$t^2 = \frac{1}{y + 1}$$
$$t = \pm\sqrt{\frac{1}{y + 1}}$$
$$x = \frac{1 - t^2}{t}$$
$$y = \frac{1 - t^2}{t^2}$$
$$\frac{x}{y} = \frac{1 - t^2}{t} \times \frac{t}{1 - t^2}$$
$$\frac{x}{y} = t \tag{3}$$
Substitute $(3)$ into $(1)$
$$x = \frac{y}{x} - \frac{x}{y}$$
$$x = \frac{y^2 - x^2}{xy}$$
$$x^2y = y^2 - x^2$$
$$x^2y + x^2 = y^2$$
$$x^2(y + 1) = y^2$$
$$x^2 = \frac{y^2}{y + 1}\tag{4}$$
$$x = \sqrt{\frac{y^2}{y + 1}}$$
From $(4):$
$$x^2y + x^2 = y^2$$
$$y^2 - x^2y = x^2$$
$$y(y-x^2) = x^2$$
I have no idea how to progress from that point. I've tried different methods in paper to make $y$ the subject of the formula, but none of them were fruitful.
How do I make $y$ the subject of the formula in $(4)$?

from your second equation we get
$$t^2=\frac{1}{1+y}$$ for $$y> -1$$ and taking the square root we have
$$t_{1,2}=\pm\frac{1}{\sqrt{1+y}}$$ thus we get
$$x=\pm\sqrt{1+y}-\pm\frac{1}{\sqrt{1+y}}$$
from the first equation we get
$$0=t^2+tx-1$$ solving this we get
$$t_{1,2}-\frac{1}{2}x\pm\sqrt{\frac{x^2}{4}-1}$$
from here you will get
$$y=\frac{1}{\left(-\frac{1}{2}x+\sqrt{\frac{x^2}{4}+1}\right)^2}-1$$
or
$$y=\frac{1}{\left(-\frac{1}{2}x-\sqrt{\frac{x^2}{4}+1}\right)^2}-1$$
