The construction:
Consider $xy=1$ revolved about $y=x$ to bring the surface, $S$ into $\Bbb R^3.$ Then rotate $S$ to get:
$$ x^2+y^2+z^2-4xy-4yz-4zx=-1 $$
Now change the coordinates of $xy=1$ and,
consider $\log(x)\log(y)=1$ revolved about $y=x$ to bring the surface $L$ into $\Bbb R^3.$ Then rotate the surface so that $L$ lies completely in the first octant $(+,+,+).$
Then take $L$ and change the coordinates s.t. every point $p=(x,y,z)$ on the surface becomes $p'=(\log(x),\log(y),\log(z)).$ Define the new surface as $L'.$
How do you show that $S\ne L'?$
The point is, that the order matters here. In one case I revolved $xy=1$ and in the other case I changed the coordinates, revolved the curve and mapped it back with the inverse change of coordinates. I thought that mapping it back would yield the surface of revolution made with $xy=1$ but apparently that's not the case?