$f^3 + g^3=1$ for two meromorphic functions

Can you find two non-constant meromorphic functions $f,g$ such that $f^3 +g^3=1$?

Yes. According to remark 2 on p.236 of Remmert's book, Classical topics in complex function theory, Springer GTM 172, a (more or less) explicit example can be given by using the Weierstrass $\wp$-function associated to the triangular lattice $\{m + n e^{2\pi i /3}\,:\,m,n \in \mathbb{Z}\}$. This function satisfies the differential equation $$(\wp')^2 = 4 \wp^3 - g_2 \wp - g_3 \qquad \text{with}\; g_2 = 0 \; \text{and} \; g_3 = \pm \frac{\Gamma\big(\frac{1}{3}\big)^{18}}{(2\pi)^6}.$$

Putting $a = \frac{1}{2} \sqrt[3]{\phantom{|}g_{3}}$ and $b = \dfrac{1}{\sqrt{24a}}$ the functions $$f = \frac{a + b\, \wp'}{\wp} \qquad \text{and} \qquad g = \frac{a - b\,\wp'}{\wp}$$ satisfy $f^3 + g^3 = 1$ by the above differential equation (the values of $a$ and $b$ are found using the ansatz $\big(a + b\,\wp'\big)^3 + \big(a - b\,\wp'\big)^3 = \wp^3$ and the differential equation).

Observe however that the zeroes of $\wp$ and the poles of $\wp'$ are poles of both $f$ and $g$ and indeed, according to the proposition on the same page this is necessarily the case (by a beautiful application of Picard's theorem). In fact, this proposition states: If $f$ and $g$ are meromorphic functions on $\mathbb{C}$ satisfying $\,f^n + g^n = 1$ for some $n \geq 3$ then $f$ and $g$ are both either constant or they have common poles. Added: In fact, as user8268 points out in his comment below, one can even show that there are actually no non-constant such functions for $n \geq 4$ (thanks a lot for that!). This is also alluded to by Remmert in his remark 1 on the same page 236.

See Remmert's book for further details.

Added Later.

Following the links on the Wikipedia page I stumbled over Hans Lundmark's beautiful page on elliptic functions where you can find some domain coloring plots of some $\wp$-functions towards the end of the page.

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 $g(z)=f(-z)$.

As I mentioned in this other post, A.C. Dixon studied in 1890 a class of elliptic functions based on inverting the Abelian integral

$$\int\frac{\mathrm dt}{\left(1-t^3\right)^{2/3}}$$

(actually, he studied a more general class of elliptic functions based on the cubic $x^3+y^3-3\alpha xy=1$, but I'll confine myself to the case $\alpha=0$ and refer the interested to Dixon's paper for the general case.)

In particular, there are the two Dixon elliptic functions $\mathrm{sm}(z,0)=\mathrm{sm}(z)$ and $\mathrm{cm}(z,0)=\mathrm{cm}(z)$ that satisfy $\mathrm{sm}^3(z)+\mathrm{cm}^3(z)=1$ and a bunch of other identities, and can be expressed in terms of Weierstrass $\wp$:

$$\mathrm{sm}(z)=\frac{6\wp\left(z;0,\frac1{27}\right)}{1-3\wp^\prime\left(z;0,\frac1{27}\right)}$$

$$\mathrm{cm}(z)=\frac{3\wp^\prime\left(z;0,\frac1{27}\right)+1}{3\wp^\prime\left(z;0,\frac1{27}\right)-1}$$

Both have the independent periods $\pi_3=B\left(\frac13,\frac13\right)$ (where $B(a,b)$ is the beta function) and $\pi_3\exp(2i\pi/3)$, and have poles at the points $z=\frac23\pi_3+k\pi_3,\;k\in\mathbb Z$ in the real line, as well as having a total of three poles and three zeros in each "fundamental hexagon". As can be surmised, these are related to the general solutions in Theo's answer, due to the fact that the Weierstrass function satisfies homogeneity relations.