Is it possible to parametrise $x^{\frac{1}{x-1}}=y^{\frac{1}{y-1}}$?

Question

I don't know if there is a process for parametrising $y^\frac{1}{y}=x^\frac{1}{x}$ and suspect it is not possible to do so. But if it is possible, is it also possible for the similar $x^{\frac{1}{x-1}}=y^{\frac{1}{y-1}}$? Furthermore, could the curve be isolated from the line $y=x$ observable in this Wolfram|Alpha plot?

Answer 1

A parameterization for $y^\frac{1}{y}=x^\frac{1}{x}$ is well-known (and not original by me).

Write it as $x^y = y^x$ and let $y = ry$. Then $x^{rx}=(rx)^x$. Taking $x$-th roots, $x^r = rx$. Dividing by $x$, $x^{r-1} - r$ or $x = r^{\frac1{r-1}}$ and $y = rx = rr^{\frac1{r-1}} = r^{1+\frac1{r-1}} = r^{\frac{r}{r-1}} $.

You might this in your other equation and see where it gets you.