Well...maybe we can.....
y=(-6/y)²-1 simpify
y = 36/y^2 - 1 multiply through by y^2
y^3 = 36 - y^2
And using the onsite solver, we have one real solution and two non-real solutions...
$${{\mathtt{y}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{36}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{2}}\\
{\mathtt{y}} = {{\mathtt{2}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{2}}\\
{\mathtt{y}} = {\mathtt{3}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{y}} = {\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{2.828\: \!427\: \!124\: \!746\: \!190\: \!1}}{i}\\
{\mathtt{y}} = {\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.828\: \!427\: \!124\: \!746\: \!190\: \!1}}{i}\\
{\mathtt{y}} = {\mathtt{3}}\\
\end{array} \right\}$$
y=(-6/y)²-1
$$\\y=\left(\frac{-6}{y}\right)^2-1\\\\
y=\frac{36}{y^2}-1\\\\
y^3=36-y^2\qquad y\ne 0\\\\
y^3+y^2-36=0\\\\$$
now I am going to use reaminder theorum to look for a factor
If y=3 then 27+9-36=0 great (y-3) is a factor and 3 is a root
$$\\(y^3+y^2-36)\div (y-3)=y^2+4y+12 \qquad $I used polynomial division to get this$\\\\
so\\
y^3+y^2-36=(y-3)(y^2+4y+12)\\\\\\
$Now find roots of $y^2+4y+12\\\\
\triangle = 16-48<0\qquad $therefore there are no real roots to this$\\\\
so\\
$The only real solution is $y=3$$
check