Solve $x^{x^x}=-1$ I do know that if you use tetration the equation would look like this.
$$^3x=-1$$
You could then theoretically use the super-root function to solve the equation, but I do not know how to use the super-root function nor do I know of any website online that could calculate that equation.
Mathematica also doesn't help.

So how would you attempt to solve it?
Also, in case anyone is interested, here is the graph for $$z = \Re[(x+iy)^{\Large (x+iy)^{(x+iy)}}]$$

And $$z = \Im[(x+iy)^{\Large (x+iy)^{(x+iy)}} ]$$

 A: I've just entered in this problem again and propose now to use the power series for the inversion 
 $ \;^3 W(x)= \text{reverse}(x \cdot \exp(x  \cdot \exp(x)))$ using the Lagrange series-inversion. You'll get a series with a very limited radius of convergence; however it seems to be finite and not really zero. But the signs of the coefficients alternate, so you can apply Euler-summation or similar tools at them.             
Then let us define $x$ as unknown and $u=\log(x)$ its logarithm, and $y=x^{x^x} = -1 $ the known value and $v=\log(y)$ its logarithm.                
Then $u = \;^3W(v)$ (in the range of convergence) and $x=\exp(u)$ .            
Using Euler-summation (of complex order and 128 terms for the partial series)  I arrive at
$\qquad u=0.762831989634  + 0.321812259776î \qquad$ and
$\qquad x=2.03425805694   + 0.678225493699î \qquad$. (In my older post I gave
$\qquad  x=2.03426954187 + 0.678025662373î \qquad$ by Newton-approximation).                  
The check gives $x^{x^x}=-0.998626839391   + 0.0000476837419237î$ which is by $0.00137 + 0.000047î$ apart.                  
I think this way is in principle viable, however one needs then better convergence-acceleration / summation tools. And possibly it is a meaningful starting-point for the classical Newton-approximation.                    
A longer treatize showing more details can be found in my webspace
A: In addition to $x=-1$, there are complex roots. These can be found numerically. For example,
fsolve(x^(x^x) = -1, complex, avoid = {x = -1});

$$- 0.1589087516+ 0.09682319092\,i  $$
A: Clearly $x=-1$ is a solution. Here I'll prove that it's the only real solution, complex solutions are a different matter.
Given $z,\alpha \in \mathbb{C}$ , we have
$$z^{\alpha} = \exp(\alpha [\log |z| + (\arg z)i])$$
So let $x = re^{i\theta} \in \mathbb{C}$. Then
$$\begin{align}
x^{x^x} &= \exp(x^x[\log r + \theta i]) \\
&= \exp\big( \exp(x\log r +\theta i)[\log r + \theta i]\big)\\
&= \exp\big( r^x(\cos \theta + i \sin \theta)(\log r + \theta i) \big)\\
&= \underbrace{\exp\big( r^x(\cos \theta \log r - \theta \sin \theta) \big)}_{\in \mathbb{R}^+} \cdot \exp \big( r^x(\sin \theta \log r + \theta\cos \theta)i \big)
\end{align}$$
Thus $\arg x^{x^x} = r^x(\sin \theta \log r + \theta \cos \theta)$.
For example, if $x \in \mathbb{R}$ and $x<0$ then $\arg x^{x^x} = -(-x)^x\pi$. So if we're going to have $x^{x^x}=-1$ and $x \in \mathbb{R}$ then certainly we need $x<0$ so that $\arg x^{x^x} = \pi$. Hence if $x<0$ and $x^{x^x}=-1$ then we have
$$-(-x)^x = -1$$
which is equivalent to $(-x)^{(-x)}=1$. As the only solution to $y^y=1$ with $y>0$ is $y=1$, this means that $x=-1$ is the only real solution.
