Find the maximum of $x^{x^{x^{⋰}}}.$ 
Question: Find the maximum of $x^{x^{x^{⋰}}}.$

Let $y = x^{x^{x^{⋰}}}.$ Then 
\begin{align}
y & = x^y \\
\Rightarrow \ln y & = y\ln x \\
\Rightarrow \frac{1}{y} \frac{dy}{dx} & = y\left(\frac{1}{x}\right) + \ln x \cdot \frac{dy}{dx}.
\end{align}
Since we are looking for maximum, we set $\frac{dy}{dx} = 0.$ So, 
$$\frac{y}{x} = 0$$
$$\Rightarrow y = 0.$$
I am not sure what's wrong here.

If we are allowed to consider values of $x$ s.t. this tends to $\infty$, then the answer is trivially $\infty$. Assuming the question is concerned with the interval over which this converges to real numbers though:
Note that when $y=0$, you get $0=x^0$, which is a contradiction. Instead, the maxima in this case occurs when $y'=\infty$. Dividing everything by $y'$ and letting it go to infinity gives us
$$\frac1y=\frac y{xy'}+\ln(x)$$
$$\frac1y=\ln(x)\tag{as $y'\to\infty$}$$
$$1=y\ln(x)$$
Since we also know that $\ln(y)=y\ln(x)$, we end up with $\ln(y)=1$, or $y=e$, which occurs at $x=\sqrt[e]e$.
For more information on convergence, see here.
