TITLE: $f^3 + g^3=1$ for two meromorphic functions
QUESTION [25 upvotes]: Can you find two non-constant meromorphic functions $f,g$ such that $f^3 +g^3=1$?

REPLY [8 votes]: As an addendum of sorts to Theo's answer, the two functions $f$ and $g$ in the answer can be simplified through the use of the homogeneity relations
$$\wp\left(cz;\frac{g_2}{c^4},\frac{g_3}{c^6}\right)=\frac1{c^2}\wp\left(z;g_2,g_3\right),\quad \wp^\prime\left(cz;\frac{g_2}{c^4},\frac{g_3}{c^6}\right)=\frac1{c^3}\wp^\prime\left(z;g_2,g_3\right)$$
Focusing on the case of positive $g_3$ (the treatment for negative $g_3$ is similar), we have
$$f(z)=\frac{3+\sqrt{3}\wp^\prime\left(\frac{\Gamma(1/3)^3}{2\pi}z;0,1\right)}{6\wp\left(\frac{\Gamma(1/3)^3}{2\pi}z;0,1\right)},\quad g(z)=\frac{3-\sqrt{3}\wp^\prime\left(\frac{\Gamma(1/3)^3}{2\pi}z;0,1\right)}{6\wp\left(\frac{\Gamma(1/3)^3}{2\pi}z;0,1\right)}$$
where now only "equianharmonic case" Weierstrass elliptic functions are involved.
If these functions are plotted in the complex plane, a hexagonal structure similar to what is observed for the Dixon elliptic functions can be seen. This suggests that there might be a relationship between these functions and the Dixon elliptic functions.
In particular, using the homogeneity relations, one can show that
$$f(z)=-\frac{\mathrm{cm}\left(\frac{\Gamma(1/3)^3}{2\pi}\sqrt{3}z\right)}{\mathrm{sm}\left(\frac{\Gamma(1/3)^3}{2\pi}\sqrt{3}z\right)}$$

As promised, here are plots of $f(z)$ on the real line:

a contour plot of the real and imaginary parts of $f(z)$:

and plots of a single "hexagonal tile" of $f(z)$:


Plots of $g(z)$ are similar since $g(z)=f(-z)$.