Solution verification: Derivative of the infinite power tower $y(x) = x^{x^⋰}$

Question

I was doing the problem $(x^{x^{⋰}})'$ and I would like someone to verify my solution: \begin{align*} &\left(y=x^{x^{^{⋰}}}\right)'\\ \implies & \;\left(y=x^{y}\right)'\\ \implies & \;\left(y^{\frac{1}{y}}=x\right)'\\ \implies & \;\left(\frac{\ln\left(y\right)}{y}=\ln\left(x\right)\right)'\\ \implies & \;\frac{y'-y'\ln\left(y\right)}{y^{2}}=\frac{1}{x}\\ \implies & \; y'-y'\ln\left(y\right)=\frac{y^{2}}{x}\\ \implies & \; y'=\frac{y^{2}}{\left(1-\ln\left(y\right)\right)x}\\ \implies & \; \left(x^{x^{⋰}}\right)'=\frac{\left(x^{x^{⋰}}\right)^{2}}{\left(1-\ln\left(x^{x^{⋰}}\right)\right)x} \end{align*}

Answer 1

May be easier would be to use Lambert function since$$ x^{x^⋰}= y \implies x^y = y\implies 1 = yx^{-y} = ye^{-y\log x}\implies y = \frac{W(-\log x)}{(-\log x)}$$ Let $t=-\log(x)$

$$\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$$ $$\frac d{dt} \Big[\frac{W(t)}t\Big]=-\frac{W(t)^2}{t^2 (W(t)+1)}$$ $$\frac{dt}{dx}=-\frac 1x$$