Prove that an entire complex-valued $f$ is constant.

If a complex-valued function $f = u + iv$ is entire with $uv = 3$ for all $z \in \mathbb C$, then $f$ is constant.

$f$ is not constant $\rightarrow f^2 = (u^2 - v^2) + 2iuv$ is not constant. Since $uv = 3$ for all $z \in \mathbb C$, $u^2 - v^2$ is not constant. The first-order partials are:

$$2u \frac{\partial u}{\partial x} - 2v \frac{\partial v}{\partial x} = 2u \frac{\partial v}{\partial y} + 2v \frac{\partial u}{\partial y} = 2 \frac{\partial}{\partial y} uv = 0$$ $$2u \frac{\partial u}{\partial y} - 2v \frac{\partial v}{\partial y} = -2u \frac{\partial v}{\partial x} - 2v \frac{\partial u}{\partial x} = -2 \frac{\partial}{\partial x} uv = 0$$

The equalities to zero result from the premise that $uv = 3$, but the first-order paritals cannot be zero simultaneously since $u^2 - v^2$ is not constant.

In my attempt, you can see that I did not use the fact that $f$ is entire. The argument seems to apply to any complex domain over which $f$ is analytic, but I already know that is not true. A counter-example is $f = 3xy + i(xy)^{-1}$ with $(x, y) \neq 0$. This function is analytic over $\mathbb C - \{0\}$ and $uv = 3$, but clearly $f$ is not constant. So, what did I do wrong, or what am I missing in an entire $f$?

Your argument is correct. Your counterexample is wrong. $f = 3xy + i(xy)^{-1}$ is not analytic. Among other things $(xy)^{-1}$ is not harmonic.