To start, I'll say that for this post I'll be using Rudy Rucker notation for tetration. That being $^2$x=$x^x%$, which means the number raised to the left means how many times one would exponentiate x.
Ok, so now to see my logic I'll start with a pattern:
$^4x$ = $x^{x^{x^x}}$ = $x^{x^{x^{x^1}}}$
$^3x$ = $x^{x^{x}}$ = $x^{x^{x^1}}$
$^2x$ = $x^x$ = $x^{x^1}$
$^1x$ = $x$ = $x^1$
Take note how for each tetration we go down by one, the previous exponent just reverts to a one, so this pattern must be true for the $0$th tetration of x.
$^0x$ = $1$ = $1^1$
Now, my question, what about for $^{-1}x$. I didn't even know if this was possible, after jut a quick scour on the internet I ended up with zilch. So I began looking for a nice formula to help. Then I came across this nice formula:
$\frac{d}{dx}$[$^nx$]=($^nx$)$\left(\frac{^{n-1}x}{x}+\frac{d}{dx}[^{n-1}x]ln(x)\right)$
Assuming n=0, and plugging in, we know from before that $^0x$=1 therefore $\frac{d}{dx}$[$^0x$]=$0$, one returns that:
$0$=$^{(n-1)}x$+$\frac{d}{dx}$[$^{(n-1)}x$] * ln$(x)$ * $x$
To solve this problem I just substituted $u=^{(n-1)}x$. So my questions arise,
Am I allowed to make this substitution?
Even if I do, do I change the variable of differentiation?
If this method I am using is not viable, is there some other method that I can use, or is it truly impossible and I just wasted a few days of my life?
It is not a matter of being "allowed" to do these things - but you have to be careful by what you mean: in particular, by the definition
$$^n x = \underbrace{x^{x^{\cdots^x}}}_\text{$n$ copies of $x$}$$
there is no assignment of $^{-1}x$ because there is no such thing as $-1$ "copies" of $x$, i.e. $^{-1} x$ is undefined in value. So when you say
$$^{-1} x = 0$$
you are actually implicitly extending the definition of $^n x$. And you certainly can do this and, I'd say, in light of all the preceding operations this definition is eminently reasonable. However, at $n = -2$, it will fail because $\log_x(0)$ does not exist. So you can only go that far, which is kinda weird, but the factorial is kind of the same way, $0! = 1$, but $(-1)!$ fails due to division by zero.