It can be shown that taking $y = \frac{1}{x}$ and rotating about the origin can produce (part of) a hyperbola (or already is a hyperbola, technically). Is there such a relation in $3$D? Does the equation $xyz = 1$ form a hyperboloid?
I tried figuring this out with a similar approach as one would in $2$D: converting to another coordinate system, subtracting an angular offset ($\phi\rightarrow\phi-\phi_0$), and converting back to Cartesian. This results in $xy\cos{(2\phi_0)} + \frac{1}{2}(y^2-x^2)\sin{(2\phi_0)}=1$, which gives $y=1/x$ for $\phi_0=0$ and $y^2-x^2=2$ for $\phi_0=\pi/4$.
Trying this with spherical coordinates, it became difficult to convert back to Cartesian without leftover terms. I have the equation $r^3\sin{(\phi-\phi_0)}\cos{(\phi-\phi_0)}\sin^2{(\theta-\theta_0)}\cos{(\theta-\theta_0)}=1$. (In both cases, I use the physics convention for coordinates where $\phi$ is used for rotations in the $XY$ plane.). I tried plotting this and offsetting the angle. $\phi$ worked as expected, but offsetting $\theta$ resulted in a weird artifact if it crossed the $Z$-axis.
The normal hyperboloids are surfaces of degree $2$ in $\mathbb R^3$. But $xyz=1$ is a surface of degree $3$. So it is not a "hyperboloid" in that sense.
I seem to recall: Newton worked out the possible surfaces of degree $3$ in $\mathbb R^3$ ?? Or did he only do curves of degree $3$ in $\mathbb R^2$ ??