Showing that $f$ is a constant function? I am working on the question:

Suppose $f$ is entire, with real and imaginary parts $u$ and $v$ satisfying $u(z) v(z) = 3$ for all $z$.
  Show that $f$ is constant.

I understand that I must show $f'(z) = 0$. The first thing I did here was to set up the Cauchy Riemann Equations: $u_x = v_y$, and $u_y = -v_x$. However, I am not sure how to use $u(z) v(z) = 3$. Clearly I have to take some sort of derivative, but I'm not very familiar with how partial derivatives work. Can I take the partial derivative of both sides of $u(z) v(z) = 3$ and say $u_x (z) v(z) + u(z) v_x(z) = 0$? If yes, how should I proceed afterwards? I have worked under this assumption, but I still didn't get the desired $f'(z) = 0$. 
 A: If $f(z)$ is entire then so is $f(z)^2$, and if $f=u+iv$ then
$$ f^2=(u+iv)^2=u^2-v^2+2iuv$$
The condition $u(z)v(z)=3$ therefore implies that $f(z)^2$ has a constant imaginary part, so now it's enough to show that an entire function with a constant imaginary part is constant.
A: Note that if $f=u+iv$ and $v=3/u$, then the Cauchy-Riemann Equations reveal that
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\implies u_x=-3u_y/u^2 \tag 1$$
and
$$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}\implies u_y=3u_x/u^2 \tag 2$$
Using $(1)$ and $(2)$ together yields
$$u_x^2+u_y^2=0\implies u=\text{constant}$$
Hence $f$ is constant.
A: Consider the function $\displaystyle g(z)=e^{if^2(z)}$. Then , $g$ is an entire function. 
Now ,  $\displaystyle |g(z)|=\left|e^{i(u^2-v^2)}\right|.\left|e^{-2uv}\right|=e^{-6}$ , for all $z\in \mathbb C$. So , $g$ is bounded in $\mathbb C$.
So by Liouville's theorem $g$ is constant and hence $f$ is constant.
A: Differentiating partially w.r.t $x$ and $y$ separately, you get $uv_x+vu_x=0$ and $uv_y+vu_y=0$. Use C-R equations to get $-uu_y+vu_x=0$ and $uu_x+vu_y=0$. On solving for $u_x$ you get 

$(u^2+v^2)u_x=0\implies u_x=0$ or $u=constant$ as $(u^2+v^2)\ne 0$ unless $u=v=0$
