Given $z^{z^x}=x$, estimate $\frac{dx}{dz}$

Consider the implicit function $$F(x,z)=z^{z^x}-x=0$$ $$\frac{\partial F(x,z)}{\partial x}=z^{z^x+x} \log ^2(z)-1$$ $$\frac{\partial F(x,z)}{\partial z}=z^{z^x} \left(z^{x-1}+x z^{x-1} \log (z)\right)$$ $$\frac {dx}{dz}=-\frac{\frac{\partial F(x,z)}{\partial z} } {\frac{\partial F(x,z)}{\partial x} }=-\frac{z^{z^x+x-1} (x \log (z)+1)}{z^{z^x+x} \log ^2(z)-1}$$

More pleasant would be the reverse way using $$z=\left(\frac{x \log (x)}{W(x \log (x))}\right)^{\frac{1}{x}}$$ then logarithmic differentiation and chain rule with $t=x \log(x)$. We should end with $$\frac {dx}{dz}=f(x)$$ which is positive for $0 \leq x \lt e$ with a vertical asymptote at $x=\frac 1e$. For $x>e$, the derivative is always negative.

Edit

For the calculation of the derivative, start with $$z=\left(\frac{t}{W(t)}\right)^{\frac{1}{x}} \qquad t=x\log(x)$$ $$\log(z)={\frac{1}{x}} \log\left(\frac{t}{W(t)}\right)$$ $$\frac {z'} z=-\frac 1{x^2}\log\left(\frac{t}{W(t)}\right)+\frac 1x \frac d {dt}\Bigg[\log\left(\frac{t}{W(t)}\right) \Bigg]\frac {dt} {dx}$$ $$\frac d {dt}\Bigg[\log\left(\frac{t}{W(t)}\right) \Bigg]=\frac{W(t)}{t (1+W(t))}$$ All of the above gives $$z'=\frac z {x^2 t (1+W(t)) }\Bigg[x W(t) t'-t (1+W(t)) \log \left(\frac{t}{W(t)}\right) \Bigg]$$ $$\frac {dx}{dz}=\frac 1 {z'}$$