The surface $x + y + z = \sin xyz$: Can it be visualized?

Question

The surface $$x + y + z = \sin xyz$$ is very hard to visualize:

What can be said about it, geometrically or visually?

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Results so far

Start with a similar, but simpler, curve: $$x + y = \sin xy.$$ It oscillates between the lines $x + y = \pm 1$. Applying the coordinate transform $$u = x - y \\ v = x + y$$ (and, for simplicity, ignoring any stretching), we get $$v = \sin \left(\frac{v^2}{4} - \frac{u^2}{4} \right) \approx - \sin \left(\frac{u^2}{4} \right)$$ which describes it quite well. This approximation is exact when $u = 0$ and works surprisingly well numerically for all values of $u$ (though it's hard for me to explain why, other than noting that $|\frac{v^2}{4}| \leq \frac 1 4$ whereas $u$ is unbounded).

Applying this to our original question: The surface is bounded by the planes $x + y + z = \pm 1$ and oscillates between them. A linear transform $(x,y,z) \to (u,v,w)$ which sends those planes to $w = k$ would simplify things. There are infinite transforms to choose. Which one would preserve symmetry or otherwise highlight the salient features? And, once we do it, how do we approximate the surface?

Remaining hurdles

How is the approximation $v = \sin \left(\frac{v^2}{4} - \frac{u^2}{4} \right) \approx - \sin \left(\frac{u^2}{4} \right)$ justified? Why does it work so well?

How can we apply it to the surface in question?

And, what is the best $u,v,w$ transform to use?

If we can solve those, we can answer the question; other approaches are, of course, equally germane.

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Update

A little solid geometry shows, perhaps surprisingly, that the positive $x$ and $y$ axes, when projected onto the plane above, have angle $\frac{2 \pi}{3}$. That suggests we should see $120^\circ$ rotational symmetry (equilateral triangle). And since the pos $x$ and neg $y$ have half that angle, we may also see $60^\circ$ rotational symmetry (equilateral hexagon).

That implies that we should use a cylindrical coordinate system, with the axis of the cylinder normal to the $x + y + z = 0$ plane, and with the $\theta = 0$ half plane at $x = y$. Using a computer, I was able to transform the surface to $$ h = \frac{1}{\sqrt 3} \sin \left( \frac{\sqrt 3}{18}(2h^3 - 3hr^2 - \sqrt 2 r^3 \cos 3 \theta) \right). $$

Assuming a similar approximation to the planar case is valid (and I'm still exploring that, we can write $$ h \approx - \frac{1}{\sqrt 3} \sin \left( \frac{\sqrt 6}{18} r^3 \cos 3 \theta) \right)$$ which visualizes quite nicely:

Answer 1

Since $\sin$ of any real number is between $-1$ and $1$, the surface lies between the planes $x+y+z=1$ and $x+y+z=-1$. For any $k \in [-1, 1]$, its intersection with $x + y + z = k$ is composed of infinitely many curves which project to $x y (k - x - y) = \theta$ in the $xy$ plane, where $\theta$ is any one of the infinitely many values such that $k = \sin(\theta)$. Here, for example, is part of that projection for $k=1$.

Answer 2

Plotted on Mathematica, surface with 3 fold symmetry at$ 120 ^\circ$ period around the normal, the normal to the surface is at center of monkey saddle with direction cosines $(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3}).$

Answer 3

Here is a rotating 3d plot with coordinates $(x,y,z)$ from $1$ to $3$.