First of all, I must admit that the topic of Tropicalization is new to me. I have a rough outline and I'm by no means asking for a ready-made solution here, but rather some pointers on how to get an initial visualization most straight-forwardly. Solution hints that require tools (or a combination of them) like Mathematica, Matlab, Maple are absolutely no problem - in fact, they are very welcome.
I have the following curve of degree six:
$$0=16y^6+48x^2y^4-20y^4+48x^4y^2-40x^2y^2+5y^2+16x^6-20x^4+5x^2-x$$
The curve is plotted by the following figure (red curve):
The blue curve is of degree eight and hopefully (based on the computational pilot for the red curve) I can then transfer the approach to the blue curve by my own.
What I understood so far is that I have to extract the coefficients of all monomials $(i,j)$ for $x^iy^j$ of our curve.
Since our curve is of degree $6$, we have $36$ coefficients, namely for monomials given by $(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 0), (2, 1), (2, 2), (2, 3), (2, 4), (3, 0), (3, 1), (3, 2), (3, 3), (4, 0), (4, 1), (4, 2), (5, 0), (5, 1), (6, 0)$
These coefficients then are: $(0, 0, 5, 0, -20, 0, 16, 0, -1, 0, 0, 0, 0, 0, 0, 5, 0, -40, 0, 48, 0, 0, 0, 0, 0, 0, -20, 0, 48, 0, 0, 0, 0, 16, 0, 0)$
Then I divided these coefficients by $48$ (the largest one) to obtain "normalized" coefficients that lie within the intervall $[-1,1]$: $(0, 0, 0.104166666, 0, -0.4166666, 0, 0.33333, 0, -0.020833333, 0, 0, 0, 0, 0, 0, 0.104166666, 0.0, -0.833333, 0, 1, 0, 0, 0, 0, 0, 0, -0.4166666, 0, 1, 0, 0, 0, 0, 0.33333, 0, 0)$
How can I obtain most straight-forwardly a tropicalization visualized? Sketching the next steps would help me a lot finishing my task.
To complete the big picture, I am adding more background information:
The blue curve has the equation:
$$0=-64y^8-256x^2y^6+112y^6-384x^4y^4+336x^2y^4-56y^4-256x^6y^2+336x^4y^2-112x^2y^2+7y^2-64x^8+112x^6-56x^4+7x^2+x$$
Both curves can be formulated by a parameter form. The red curve within the domain $\frac{4\pi}{3}<t<2\pi$:
\begin{array}{l} x^2+y^2=y\sin t+x\cos t\\ x^2+y^2=y\sin\frac{3t}{2}+x\cos\frac{3t}{2} \end{array}
and the blue curve within its domain $0<t<\frac{4\pi}{3}$:
\begin{array}{l} x^2+y^2=y\sin t+x\cos t\\ x^2+y^2=-y\sin\frac{3t}{4}-x\cos\frac{3t}{4} \end{array}
It would intersting to know, whether we can restrict the tropicalization to these domains.