TITLE: Simultaneous zero set of two equations in $\mathbb R^3$
QUESTION [15 upvotes]: Can we have positive reals $x,y,z$ with
$$  x^{\left( y^z \right)}  =   y^{\left( z^x \right)}  =  z^{\left( x^y \right)}      $$ in cyclic permutation, other than the line $x=y=z$? 
I put this at https://math.stackexchange.com/questions/493739/this-is-stupid-but-i-have-a-bad-cold-with-cough and one poor guy has been hacking away at it. I still have no idea.
I would love to see some representation of the supposed surface $$  x^{\left( y^z \right)}  =   y^{\left( z^x \right)},      $$ which I feel really ought to be some ugly variant of the helicoid around the line $x=y=z$.
EDIT, Thursday:  Is it true that the gradient of $$ \color{magenta}{ h(x,y,z) = x^{\left( y^z \right)}  -   y^{\left( z^x \right)}},      $$ is defined and a nonzero vector along the line, say, at the point $x=y=z=t$ for positive real $t?$ If so, the surface is orthogonal to that at the line...Further, if you switch to one of the other pairs as a difference, I expect the gradient vector to be rotated by $120^\circ;$ this gives a pretty good reason for there being no other points on all three such surfaces near the known line.

REPLY [8 votes]: Maple found closed form for $z$ and parametrization over $\mathbb{C}$
for $x^{(y^z)}=y^{(z^x)}$:
$$ z= {{\rm e}^{ \left( -x{\it LambertW} \left( -\ln  \left( y \right) { {\rm e}^{\ln  \left( {\frac {\ln  \left( x \right) }{\ln  \left( y \right) }} \right) {x}^{-1}}}{x}^{-1} \right) +\ln  \left( {\frac { \ln  \left( x \right) }{\ln  \left( y \right) }} \right)  \right) {x}^ {-1}}} \qquad (1) $$ 
EDIT
Substituting $z$ in the other equation and solving, we are looking
for positive real solutions of
x^exp(ln(y)*exp((-x*LambertW(-ln(y)/x*exp(ln(ln(x)/ln(y))/x))+ln(ln(x)/ln(y)))/x))-exp((-x*LambertW(-ln(y)/x*exp(ln(ln(x)/ln(y))/x))+ln(ln(x)/ln(y)))/x)^(x^y)

Complex solutions exist, e.g.
x1= (4.039760831390928810096443230122786460497 - 3.794516696260006267408154939005426227609j)
y1= (2.764005926356850384451036006794311106205 + 0.0j)
z1= (1.273990994985239416952018723391988170796 + 0.2830754921068781180566923204291057573632j)

Things get complicated by the fact that with the principal branches
of $W$ and $\log$ $(1)$ need not be real and might have imaginary part.
Here is a plot of $(1)$ taking the principal branches and the real
part. The plot is wrong because it has some artifacts caused by 
imaginary part.
Will try to make correct plot if I have the time.