Wow! Very nice klein bottle. Not lame anymore. Please change the title. 

Here is a degree 6 formula :

((x^2+y^2+z^2)+2*y-1)*(((x^2+y^2+z^2)-2*y-1)^2-8*z^2)+16*x*z*((x^2+y^2+z^2)-2*y-1);

You can play with the numerical parameters. It's remarquable that doing so won't change the topology of the surface. It's in the structure of the polynomial not its parameters.

(I don't remember were I found the formula. That was some time ago.)

I would prefare some darker glass for Are you using fragmentarium?

I would like to discuss also my method for finding a DE, without derivatives (that's the method that Knighty followed in its script)

Too bad, it is a bit too complicated for some surfaces, because it's needed to know the signs of every "polynomial component" of the surface.

Let's examine a very very simple example.

Let's define;
x2 = x*x;
y2 = y*y + z*z;
And examine the "lathed" Scarabaeus curve expression;

(x2 + y2) (x2 + y2 + R1 x)^2 - R2^2 (x2 - y2)^2 = 0

The problem for DE is that the field needs to be "approximately linear", on x,y,z. For example a sphere equation is;

x*x + y*y + z*z - R*R = 0

To make it "linear", we can take square root. It does not exist for negative values? Oh it's simple, bring R*R to the right.

x*x + y*y + z*z = R*R
Then;
sqrt(x*x + y*y + z*z) - abs(R) = 0
This gives a very good sphere DE.

Okay, applying the same method to the scarab curve is not so plain; (x2 + y2) (x2 + y2 + R1 x)^2 is always positive (a sum of squares) but (x2 - y2)^2 has a variable sign. If we expand we have (x2 - y2)^2 = (x^4 + y^4) - 2*(x*x*y*y)
Again we can separate positive and negative, because x*x*y*y is always >0 ;

ssqrt((x2 + y2) (x2 + y2 + R1 x)^2 + 2*R2*R2* (x*x*y*y))) - sqrt(R2)*ssqrt(x^4+y^4) where ssqrt(a) is sqrt(sqrt(a)) to make the vector field of power 1 because the highest power is 4.

That seems to works like a charm!

The problem is that many surfaces have lots mixed terms like x*y, or even power terms like x, x^3 ... , so separate those correctly is a trouble.

In particular Enneper implicit equation...





 

For me your formula is wonderful but calculates three+1 times the function and it's hard to code (and slower than taking some roots). So wondering about a workaround. Later.

You are right. It's should be faster without using derivatives.