How to find $y^{(y^2-6)}$? $$\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = y$$ $$x≠2$$ $$y^{(y^2-6)} = ?$$What is $y^{(y^2-6)}$? Could you please explain to me how to solve this question step by step?

Note that
$$y=\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = \frac{3}{1-3^{x-2}} + \frac{3}{1-\frac1{3^{x-2}}}=\\= \frac{3}{1-3^{x-2}} - \frac{3^{x-1} }{1-3^{x-2}}=\frac{3-3^{x-1}}{1-3^{x-2}}=3\frac{1-3^{x-2}}{1-3^{x-2}}=3 $$
