**NB: I have edited this question to clarify what the OP is asking – Robert Bryant**


Problem: Find a holomorphic function $f$ where where $f(x+iy) = u(x,y) + i\,v(x,y)$, such that the graph $\Gamma\_u = \left\{\bigl(x,y,u(x,y)\bigr)\mid (x,y)\in\mathrm{dom}(f)\right\}\subset\mathbb{R}^3$ has Gauss curvature $K=-1$ at every point. 


Because the Gauss curvature of such a graph $\Gamma\_u$ is given by the formula
$$ 
K(x,y) = \frac{u\_{xx}u\_{yy} - u\_{xy}^2}{ (1 + u\_x^2 + u\_y^2)^2}, 
$$
the equation $K(x,y) = -1$ is a PDE of Monge-Ampere type. However, in this case, because $u$ is the real part of a holomorphic function and hence is harmonic, the function $u$ must also satisfy $u\_{xx}+u\_{yy} = 0$.


One wants to solve this combined system or else find a way to reduce it to an ODE so that one can investigate numerical solutions.



Monge-Ampere type PDE
Now I understand the OP's question, which I would state this way: 


Is there a holomorphic function $f(x,y) = u(x,y) + \mathrm{i}\,v(x,y)$ on a domain in the $xy$-plane such that the induced metric on the graph of $u$ has Gauss curvature $-1$? (Note: The OP needs $f$ to be holomorphic, not just complex, if the given formula for the Gauss curvature is going to be correct.)


The answer to this question is 'no'; here is why: The hypotheses can be expressed in terms of $u$ as follows: Find the common (local) solutions to the pair of equations
$$
u\_{xx} + u\_{yy} = 0\qquad\text{and}\qquad 
{u\_{xx}}^2 + {u\_{xy}}^2 = \bigl(1 + {u\_x}^2+{u\_y}^2\bigr)^2.
$$
I claim that there are no such solutions. To see this, note that for any solution of these equations on a simply connected domain in the $xy$-plane, there must exist a function $\theta$ on that domain such that
$$
u\_{xx} = \cos\theta\,\bigl(1 + {u\_x}^2+{u\_y}^2\bigr),\quad
u\_{xy} = \sin\theta\,\bigl(1 + {u\_x}^2+{u\_y}^2\bigr),\quad
u\_{yy} = -\cos\theta\,\bigl(1 + {u\_x}^2+{u\_y}^2\bigr).
$$
Differentiating these equations and expanding out $(u\_{xx})\_y = (u\_{xy})\_x$ and $(u\_{xy})\_y = (u\_{yy})\_x$ then shows that the function $\theta$ must satisfy the differential equations
$$
\theta\_x = 2(u\_x\,\sin\theta -u\_y\,\cos\theta),\quad\text{and}\quad
\theta\_y = -2(u\_y\,\sin\theta +u\_x\,\cos\theta).
$$
Finally, using these formulae and the above formulae for $u\_{xx}, u\_{xy}, u\_{yy}$ to expand out the identity $(\theta\_x)\_y-(\theta\_y)\_x = 0$ then yields $4=0$, a contradiction.


Thus, there are no solutions of this kind.


