What is too complicated about $y=x^{xy}$? While playing with Desmos today, I typed the equation $y=x^{xy}$ and the graph came out to be 
I clicked the Learn More option given near my equation and Desmos said: Sometimes the calculator detects that an equation is too complicated to plot perfectly in a reasonable amount of time. When this happens, the equation is plotted at lower resolution.
What is the complication?
 A: We have
$$\begin{align}&y=(x^x)^y\\
\implies &y^{\frac 1y}=x^x\\ \implies &y^{-1}\ln y=x\ln x\\
\implies &-y^{-1}\ln y^{-1}=x\ln x\\
\implies &y^{-1}\ln y^{-1}=-x\ln x\\ 
\implies &\ln y^{-1} e^{\ln {y^{-1}}}=-x\ln x\\
\implies &W\left(\ln y^{-1} e^{\ln {y^{-1}}}\right)=W\left(-x\ln x\right)\\
\implies &\ln y^{-1}=W\left(-x\ln x \right)\\
\implies &y^{-1}=e^{W\left(-x\ln x\right)}\\
\implies &y=e^{-W\left(-x\ln x\right)}\end{align}$$
This implies, your function is a non-elementary function and can be written as
$$f(x)=e^{-W\left(-x\ln x\right)}.$$

Then, note that $W(x)$ is real for only $x≥-\frac 1e$.
This means , we have
$$\begin{align}-x&\ln x≥-\frac 1e,\thinspace x>0\\
\implies &x\ln x≤\frac 1e \\
\implies &\ln x e^{\ln x}≤\frac 1e\\
\implies &W\left(\ln xe^{\ln x}\right)≤W\left(\frac 1e\right)\\
\implies &\ln x≤W\left(\frac 1e\right)\\
\implies &0<x≤e^{W\left(1/e\right)}≈1.3211\end{align}$$.
A: The solution is quite simple in terms of Lambert function
$$y=x^{xy}\implies y=-\frac{W(-x \log (x))}{x \log (x)}$$ In order to stay in the real domain, you need that
$$x \log(x) \geq \frac 1e \implies x \lt \frac{1}{e W\left(\frac{1}{e}\right)}=1.32110\cdots$$
We also have
$$y'=\frac{ e^{-2 W(-x \log (x))}}{W(-x \log (x))+1}(\log (x)+1)$$ which cancels at $x=\frac 1e$ and at this point $y=e W\left(\frac{1}{e}\right)=0.756945\cdots$
I think that we have a good description of your function.
