Expanding on some of this...

I suppose it makes no sense unless $x$ is between 0 and 1. And in this case...

Define $y =x^{x^{x^{\cdots}}}$. Then we have $y= x^y$. For any fixed $0<x<1$, this defines a unique positive value for $y$ (intermediate value theorem will imply $y$ is between 0 and 1).

So that's one way to define it in the range we want. It seems to me that solving the above equation gives us:

$$x = e^{\log(y)/y} = y^{1/y}$$

or

$$ 1/x = (1/y) ^{1/y}.$$

This can be written in terms of the Lambert W function to give:

$$y = e^{-W(-\log(x))}.$$

I don't suppose we could get more explicit than that.

But your situation is much easier since you already know $y$. So your question is the same as "is $y= \zeta(2)-1$ a solution to $y^{1/y} = 1/2$." (No.)

Expanding on some of this...

I suppose it makes no sense unless $x$ is between 0 and 1. And in this case...

Define $y =x^{x^{x^{\cdots}}}$. Then we have $y= x^y$. For any fixed $0<x<1$, this defines a unique positive value for $y$ (intermediate value theorem will imply $y$ is between 0 and 1).

So that's one way to define it in the range we want. It seems to me that solving the above equation gives us:

$$x = e^{\log(y)/y} = y^{1/y}$$

or

$$ 1/x = (1/y) ^{1/y}.$$

This can be written in terms of the Lambert W function to give:

$$y = e^{-W(-\log(x))} = \dfrac{W(-\log(x))}{-\log(x)}.$$

We won't get more explicit than that.

But your situation is much easier since you already know $y$. So your question is the same as "is $y= \zeta(2)-1$ a solution to $y^{1/y} = 1/2$." (No.)

Expanding on some of this...

I suppose it makes no sense unless $x$ is between 0 and 1. And in this case...

Define $y =x^{x^{x^{\cdots}}}$. Then we have $y= x^y$. For any fixed $0<x<1$, this defines a unique positive value for $y$ (intermediate value theorem will imply $y$ is between 0 and 1).

So that's one way to define it in the range we want. It seems to me that solving the above equation gives us:

$$x = e^{\log(y)/y} = y^{1/y}$$.

So your function is the inverse of this. Might be related to the Lambert W function (didn't think).

Expanding on some of this...

I suppose it makes no sense unless $x$ is between 0 and 1. And in this case...

Define $y =x^{x^{x^{\cdots}}}$. Then we have $y= x^y$. For any fixed $0<x<1$, this defines a unique positive value for $y$ (intermediate value theorem will imply $y$ is between 0 and 1).

So that's one way to define it in the range we want. It seems to me that solving the above equation gives us:

$$x = e^{\log(y)/y} = y^{1/y}$$

or

$$ 1/x = (1/y) ^{1/y}.$$

This can be written in terms of the Lambert W function to give:

$$y = e^{-W(-\log(x))}.$$

I don't suppose we could get more explicit than that.

But your situation is much easier since you already know $y$. So your question is the same as "is $y= \zeta(2)-1$ a solution to $y^{1/y} = 1/2$." (No.)