How to define a surface which includes three coordinate axes? Imagine a plane with two crossing lines along $x$ and $y$ axes, and one line $y=x$. Now bend the $y=x$ line towards $z$ axis, dragging the part of that plane with it. When this line coincides with $z$ axis, you'll get a surface similar to this:

But this one was generated from $z=\text{gaussian}_{0.3a}(x)$, i.e. if continued it'd not have $z$ full axis as its subset.
I'd like the surface to be:


*

*Differentiable (except maybe at origin)

*Symmetric in the sense that for each axis $A$ there existed a rotation which placed another axis $B$ to original location of $A$, leaving the surface unchanged.

*Appeared as a single bent & stretched plane


How could I find (analytically define) such a surface (or a family of surfaces)?

Consider the monkey saddle, equation $z = x^3 - 3 xy^2 $. Note that this graph has 6-fold rotational symmetry, and along the lines $y=0$, $y= \frac{\sqrt 3} 2 x$, and $y = \frac{ - \sqrt 3} 2 x$, increases like $z = r^3$. Hence, taking the cube root, $z = \sqrt[3]{x^3 - 3 xy^2}$ gives us a function which is linear along 3 equally spaced directions.
Compare this with the coordinate $x,y,z$-axes when viewed from the point $(1,1,1)$; this also looks linear along 3 equally spaced directions, but with slope $1/2$ in these directions. Hence it follows that the function
$ z = \dfrac 12 \cdot \sqrt[3]{x^3 - 3xy^2},$
rotated appropriately, is your desired function.
