I was able to parametrize $\ln(x)\ln(y)=1$ with $x=e^{t}$ and $y=e^{1/t}.$
Now I'm interested in a parametrization and cartesian equation for the surface of revolution for $\ln(x)\ln(y)=1$ about the line $y=x.$
I tried using wolfram alpha but it gave me a very complicated parametrization. I also could not find the cartesian equation from it. Looking for something simpler.
On
It is sufficient to consider half the curve, parametrized by $$\sigma:\quad t\mapsto\left\{\eqalign {x(t)&:=e^t \cr y(t)&:=e^{1/t}\cr}\right.\qquad(1\leq t<\infty)\ .\tag{1}$$ Keep $t\geq1$ fixed for the moment. Under the desired rotation the curve point $\sigma(t)$ moves on a circle $\gamma$ with center $m=(x_m,y_m,0)$ on a plane orthogonal to the line $y=x, \> z=0$. It is easy to see that $$(x_m,y_m,z_m)=\left({x(t)+y(t)\over2}, \>{x(t)+y(t)\over2}, \>0\right)\ ,\tag{2}$$ and that the radius $r$ of $\gamma$ is given by $$r=\sqrt{2}\>{x(t)-y(t)\over2}\ .\tag{3}$$ We now need a parametrization of $\gamma$. To this end we introduce the rotation angle $\phi$ as parameter. We obtain $$\gamma:\quad\phi\mapsto\left\{\eqalign{ \bar x(\phi)&=x_m+{1\over\sqrt{2}}r\cos\phi\cr \bar y(\phi)&=y_m-{1\over\sqrt{2}}r\cos\phi\cr \bar z(\phi)&=r\sin\phi \cr}\right.\qquad(0\leq\phi\leq2\pi)\ .$$ Replace here $x_m$, $y_m$ and $r$ by the values (depending on $t$) given in $(2)$ and $(3)$, and you obtain a parametrization of your rotational surface $S$: $$S:\quad (t,\phi)\mapsto \left\{\eqalign{ X(t,\phi)&={x(t)+y(t)\over2}+{x(t)-y(t)\over2}\cos\phi\cr Y(t,\phi)&={x(t)+y(t)\over2}-{x(t)-y(t)\over2}\cos\phi\cr Z(t,\phi)&={x(t)-y(t)\over\sqrt{2}}\sin\phi \cr}\right.\quad\qquad(t\geq1, \ 0\leq\phi\leq2\pi)\ .$$ Finally plug in $x(t)$, $y(t)$ according to $(1)$.
Let us first look at an easier example.
Suppose that we are given a curve of the form $r(t) = x(t)\hat{i} + y(t)\hat{j}$ and we want to rotate it about the $x-$axis. That is, about the line passing through the origin in the direction of $\hat{i}.$
It is not too difficult to prove it that the surface of revolution will be parameterised as $S(t, \theta) = x(t)\hat{i} + y(t)\cos\theta\hat{j} + z(t)\sin\theta\hat{k},$ where $\theta \in [0, 2\pi)$ and $t$ has its original domain.
Now, if we take a step back and observe, what really mattered was that $\{\hat{i}, \hat{j}, \hat{k}\}$ was an orthonormal set. That is, given any vector of the set, it was of unit length and moreover, any two vectors were perpendicular. Moreover, the set $\{\hat{i}, \hat{j}\}$ ``spanned'' the original region of the curve, that is, the $XY-$plane.
To answer your question, we need to rotate the curve about $y = x.$ That is, about the line passing through the origin in the direction of $\frac{1}{\sqrt{2}}(\hat{i} + \hat{j}).$ This will ''play the role'' of $\hat{i}$ here.
The ``role'' of $\hat{j}$ will now be played by $\frac{1}{\sqrt{2}}(\hat{i} - \hat{j}).$
$\hat{k}$ will remain as before. It can be checked that the set $\left\{\frac{1}{\sqrt{2}}(\hat{i} + \hat{j}), \frac{1}{\sqrt{2}}(\hat{i} - \hat{j}), \hat{k}\right\}$ has the same desired properties as before.
Now, we rewrite the given curve $r(t) = x(t)\hat{i} + y(t)\hat{j}$ as:
$r(t) = \frac{1}{\sqrt{2}}(x(t) + y(t))\left[\frac{1}{\sqrt{2}}(\hat{i} + \hat{j})\right] + \frac{1}{\sqrt{2}}(x(t) - y(t))\left[\frac{1}{\sqrt{2}}(\hat{i} - \hat{j})\right].$
Following the same pattern as before, we get the parametric equation of the curve as:
$S(t, \theta) = \frac{1}{\sqrt{2}}(x(t) + y(t))\left[\frac{1}{\sqrt{2}}(\hat{i} + \hat{j})\right] + \frac{1}{\sqrt{2}}(x(t) - y(t))(\cos\theta)\left[\frac{1}{\sqrt{2}}(\hat{i} - \hat{j})\right] + \frac{1}{\sqrt{2}}(x(t) - y(t))(\sin\theta)\hat{k}.$
Now, we just collect back the terms and get the coefficients of the unit vectors to get our final parameterisation as:
$$X(t, \theta) = \frac{1}{2}\left[x(t)(1 + \cos\theta) + y(t)(1 - \cos\theta)\right]$$ $$Y(t, \theta) = \frac{1}{2}\left[x(t)(1 - \cos\theta) + y(t)(1 + \cos\theta)\right]$$ $$Z(t, \theta) = \frac{1}{\sqrt{2}}(x(t) - y(t))(\sin\theta)$$
In your case, you had already found $x(t) = e^t$ and $y(t) = e^{1/t},$ so that completes it.
Note: The above can be simplified further using $1 + \cos(2x) = 2\cos^2 x$ and $1 - \cos(2x) = 2\sin^2 x$.
EDIT:
Getting the cartesian equation:
Let us - once again - first look at an easier example.
Suppose that you are given a curve in the $XY-$plane given by $f(x, y) = 0.$
It can be shown easily that rotating this curve about the $x-$axis gives the surface $f(x, \sqrt{y^2 + z^2}) = 0.$ Note that the $\sqrt{y^2 + z^2}$ is the distance of a point $(x, y, z)$ from the $x-$axis.
Now, to solve your question, I need to revolve the curve about $x = y.$ However, I will not do this directly but rather, first convert to a question involving rotating the curve about the $x-$axis. It is easy to show that the transformation: $$x \mapsto \frac{1}{\sqrt{2}}(X + Y) \qquad y \mapsto \frac{1}{\sqrt{2}}(X - Y)$$ rotates the axes anticlockwise by $\pi/4$ radians (or $45^o$).
What this means is that $\ln\left(\frac{X+Y}{\sqrt{2}}\right)\ln\left(\frac{X-Y}{\sqrt{2}}\right) = 1$ should now be rotated about the $X-$axis to get the equation of the desired surface (in the new system).
This gives us the following: $$\ln\left(\frac{X+\sqrt{Y^2 + z^2}}{\sqrt{2}}\right)\ln\left(\frac{X-\sqrt{Y^2 + z^2}}{\sqrt{2}}\right) = 1$$
Now, we can simply return to our old coordinate system by the substitutions: $$X \mapsto \frac{1}{\sqrt{2}}(x - y) \qquad Y \mapsto \frac{1}{\sqrt{2}}(x + y)$$
I'll let you do the substitution for this one. :)