The region to the left and below the hyperbola $xy=1$ is to be mapped conformally and one-to-one onto the upper the half-plane. (See picture below.)
Ideas: The only thing I've seen before in terms of conformal maps and hyperbolae is that the area to the right of the hyperbola $x^2-y^2=1$ is mapped to $\text{Re}(z)>1$ by the map $z\mapsto z^2$. Here we have zero in the region, so the squaring map is not conformal there.
Any ideas?
Update: There is a nearly identical problem in Ahlfors, exercise 3.4.2 #8, as @bryanj points out. The problem there reads
Map the part of the $z$-pland to the left of the right-hand branch of the hyperbola $x^2-y^2=1$ on a half plane. Hint: Consider on one side the mapping of the upper half of the region by $w=z^2$, on the other side the mapping of a quadrant by $w=z^3-3z.$
I don't quite understand the hint. Perhaps @bryanj can clarify, since he appears to be following it.
I believe I understand Ahlfors' description of the map $z\mapsto z^3-3z$...he parameterizes regions in the $z$- and $w$-planes by a parameter $\zeta$ s.t. $z=\zeta + \frac1\zeta$ and $w=\zeta^3 + \frac{1}{\zeta^3}$.
Halving a symmetric region
When the region to be mapped has a line of symmetry $L$, consider dividing it along this line. If you map one half of the region to a quadrant so that $L$ goes to one of two sides, then reflection across $L$ produces a conformal map of the entire region onto half-plane.
In this example, the line of symmetry is $x=y$. Let's take the part to the right of this line, where $x>y$. The squaring map $w=z^2$ is conformal here. The half-hyperbola goes to the half-line $\operatorname{Im}w=2$, $\operatorname{Re}w\ge 0$. The line $L$ gets wrapped around at point $0$, and ends up being a half-line, going from $(1+i)^2 = 2i$ to $0^2=0$ then back up along the imaginary axis.
Thus, the image of half-region is bounded by he vertical half-line $[0,i\infty)$ and the horizontal half-line $[2i,2i+\infty)$. How to map this ugly thing $\Omega$ to a quadrant?
Weird trick
It turns out to be easier to do the opposite: map the positive quadrant to this ugly thing. Specifically, the map $\Phi(z) = z^3-3z$ does the following:
Indeed, $iy \mapsto -i(y+y^3)$, so the upper imaginary half-axis goes to the bottom imaginary half-axis in a monotone way. Along the real axis, $\Phi$ sends $0$ to $0$, then goes to the left until $\Phi(1)=-2$ ($1$ is its critical point), the goes to the right into $+\infty$.
The fact that $\Phi$ is injective in the open quadrant can be inferred from the boundary correspondence. A quarter-arc $|z|=R$, with $R$ large, is mapped approximately onto $3/4$ arc $|z|=R^3$. Combining this with the earlier observations about the boundary behavior of $\Phi$, conclude that the winding number of $\Phi$ about every point in its image is $1$.
Endgame
Recall our ugly thing $\Omega$: it is not exactly the image of positive quadrant under $\Phi$, but is related by a linear transformation. Specifically, $w\mapsto -iw-2$ maps $\Omega$ onto the image of $\Phi$. Hence, $$z\mapsto \Phi^{-1}(-iz^2-2)\tag{1}$$ sends half-domain onto first quadrant. Note that the line $L$ is sent to the positive real half-axis. Reflecting along $L$, we conclude that (1) is a map of the original region onto the right half-plane. Multiply by $i$.
This is not a reasonable exam problem, in my opinion.