I just reverse-engineered this

Note: I hope someone would be able to give a more in-depth solution than mine, as I just reverse engineered a particular function I want to play along with and came up with this problem. :) (I actually started with the solution, then created the differential equation from it. It turns out the solving the equation requires the knowledge of the Lambert W Function, which is somehow a "just learned the surface of it" concept to me at the moment.)

So, I guess I'll just show how I arrived at the said differential equation from the original function, which, by the way, is the power tower function.

$y = x^{x^{x^{x^{x^{...}}}}}$

This function gives finite values of $y$ over the interval $(e^{-e}, e^{\frac{1}{e}}]$ .

We can simplify this function to

$y = x^y$

taking logarithms of both sides,

$ln \: y = y \: ln \: x$

and by implicit differentiation,

$\frac {y'}{y} = \frac{y}{x} + (ln \: x)y'$

by separating $y'$ ,

$y' ( \frac{1}{y} - ln \: x ) = \frac {y}{x}$

so we get

$y' = \frac {y^2}{x[ 1 - y\: ln \: x ]}$

but we know that $y\:ln\:x = ln \: y$ , and $x = y^{\frac{1}{y}}$ , so the RHS becomes free of x.

That's how I just got

$y' = \frac {y^{2 - \frac{1}{y}}}{1 - ln \: y}$

Okay so I have just verified that indeed, the power tower is a solution of this differential equation. What's the significance of $(1,1)$ then?

In the event a solution has been found by solving the differential equation, an arbitrary constant will appear. Substituting that point to the values of $x$ and $y$ will determine its value and thus determine the specific function are looking for.

So using the equation $x = y^{\frac{1}{y}}$ , we can now substitute $y = e^{\frac{1}{e}}$ .

$\large x = (e^{\frac{1}{e}})^{\frac{1}{e^{\frac{1}{e}}}}$

$\large x = (e^{\frac{1}{e}})^{(e^{-\frac{1}{e}})}$

$\large x = (e^{\frac{1}{e} \cdot e^{-\frac{1}{e}} } )$

$\large x = e^{\frac{e^{-\frac{1}{e}}}{e}}$

$\large x = e^{ e^{ -( \frac{1}{e} +1)}}$

$x \approx 1.29000$

$\begin{aligned} \frac {dy}{dx} & = \frac {y^{2-\frac 1y}}{1-\ln y} \\ dx & = \frac {1-\ln y}{y^{2-\frac 1y}} dy & \small \color{#3D99F6}{\text{Integrate both sides}} \\ x & = \int \frac {1-\ln y}{y^{2-\frac 1y}} dy \\ & = \int \color{#3D99F6}{\frac {y^\frac 1y (1-\ln y)}{y^2}} dy \end{aligned}$

Note that

$\begin{aligned} \frac d{dy} y^{\frac 1y} & = \frac d{dy} e^{\frac 1y \ln y} & \small \color{#3D99F6}{\text{By chain rule}} \\ & = e^{\frac 1y \ln y} \cdot \frac d{dy} \left(\frac {\ln y}y \right) \\ & = y^{\frac 1y} \left(- \frac {\ln y}{y^2} + \frac 1{y^2} \right) \\ & = \color{#3D99F6}{\frac {y^\frac 1y (1-\ln y)}{y^2}} \end{aligned}$

Therefore, we have:

$\begin{aligned} \implies x & = \int \color{#3D99F6}{\frac {y^\frac 1y (1-\ln y)}{y^2}} dy \\ & = y^{\frac 1y} + \color{#3D99F6}{C} & \small \color{#3D99F6}{C \text{ is the constant of integration.}} \\ \implies \color{#3D99F6}{1} & = \color{#3D99F6}{1} + C & \small \color{#3D99F6}{\text{It is given that when }x=1, \ y =1} \\ \implies C & = 0 \\ \implies x & = y^{\frac 1y} = \left(e^{1/e}\right)^\frac 1{e^{1/e}} = 1.290005369 \end{aligned}$

$\implies \lfloor 1000x \rfloor = \boxed{1290}$

$y^{ \dfrac{1}{y} -2} (1- lny) dy=dx$

$y^ \dfrac{1}{y} (\dfrac{1-lny}{y^2}) dy=dx$

$\int y^ \dfrac{1}{y} (\dfrac{1-lny}{y^2}) dy= \int dx$

$y^ \dfrac{1}{y}=x+c$ where $c$ is an integration constant

From $x=1,y=1$ we found $c=0$

So the function is $x=y^ \dfrac{1}{y}$

We substiute (y)^(1/y = p and the thing on Right hand side is its derivative ie dp

Now it becomes a child's play to integrate