Does existence of non-trivial solution of $S(x,y,z) = 0, \; S(y,z,x) = 0, \; S(z,x,y) = 0$ implies existence of trivial solution at $x=y=z$ axis? My question is following. Suppose that you have an implicit surface given by equation $S(x,y,z) = 0$ (if it matters, now $S(x,y,z)$ is a polynomial function). I'm interested only in $\mathbb{R}^3_{+}$ part of $\mathbb{R}^3$ (you may think that $S(x,y,z)$ is symmetric under $(x,y,z) \mapsto (\pm x, \pm y, \pm z)$ ). 
In my case the intersection of $S(x,y,z)$ with coordinate planes will look like on the picture:

Let the cyclic group of 3rd order act on $\mathbb{R}^3$ via cyclic shift $(x,y,z) \mapsto (y,z,x)$. 

It's easy to see that if $\lbrace S(x,y,z) = 0 \rbrace \cap \mathbb{R}^3_+$ is fully contained in some fundamental domain of group action then system of equations $S(x,y,z) = 0, \; S(y,z,x) = 0 \; S(z,x,y) = 0 $ has no solution. If $S(x,y,z)$ intersects with only one face of fundamental domain, then the same result holds. 
Now suppose that we have a non-trivial solution of system $S(x,y,z) = 0, \; S(y,z,x) = 0 \; S(z,x,y) = 0 $. By non-trivial I mean solution that is inner point of group action's fundamental domain. Now the question goes here:
My hypothesis is that such surface $S(x,y,z)$ necessarily will have intersection with line $x=y=z$. Are there any counterexamples that immediately jump to your mind or is there any proof sketch?
UPDATE So, the most important feature of $S(x,y,z)$ is that $S(x,y,0) = 0$ has solution. Solution set consists of two compact connectivity components (see the picture 1 for example). The same for $S(x,0,z)$ and $S(0,y,z)$
 A: What about 
$$
S(x,y,z)=4-x^2(y^2+z^2)+yz^2(7x-y)
$$
We have $S(1,2,0)=S(2,0,1)=S(0,1,2)=0$ but $S(t,t,t)=4(t^4+1)\neq 0$
A: How about the following: Consider a torus
$$\bigl(\sqrt{\xi^2+\eta^2}-a\bigr)^2 +\zeta^2=b^2,\qquad 0<b<a,$$
in $(\xi,\eta,\zeta)$-coordinates. After a suitable squaring this can be written in the form
$$\Phi(\xi,\eta,\zeta)=0\ ,\tag{1}$$ where $\Phi$ is some polynomial of degree $4$ in the variables $\xi$, $\eta$, $\zeta$.
Then let the $\zeta$-axis point in direction $(1,1,1)$ of $(x,y,z)$-space and choose the $\xi$- and the $\eta$-axis orthogonal in the plane $x+y+z=0$. The greek and the latin variables will then be related by an orthogonal linear transformation. It follows that the torus is given in the latin variables by an equation of the form
$$\Psi(x,y,z)=0\ ,$$
where  $\Psi$ is a symmetric polynomial of degree $4$ in the latin variables. Choosing the parameters $a$ and $b$ appropriately you can make sure that the torus intersects the coordinate planes. But it will not intersect the line $x=y=z$.
If symmetry is important, here is another idea: Consider the edge-skeleton of the octahedron $|x|+|y|+|z|\leq1$ and replace the edges by tubes of radius $r\ll1$ soldered together in a smooth way. The resultant surface has the symmetries you want and does not intersect the line $x=y=z$. Some people will even find a polynomial equation $\Omega(x,y,z)=0$ that defines such a surface.
