I wonder how to give a complete metric on $\mathbb{C}\backslash \{ 0,1\} $ with Gaussian curvature $K \leqslant - 1$ explicitly.
I tried to modify the Poincaré disk model since it has constant curvature $-1$ under the hyperbolic metric. However, I still could not see how to make it right.
Is it possible to use cutoff functions to give an explicit expression?
Any help will be appreciated.
You can get an explicit formula with $K=-1$ on the nose, if you wish. Here's an outline.
There exists a holomorphic universal covering map $f : \mathbb{H} \to \mathbb{C} \setminus \{0,1\}$ which is invariant under a certain index 6 subgroup $\Gamma < \text{SL}_2(\mathbb{R})$ acting by isometries of the hyperbolic metric $\frac{\sqrt{dx^2+dy^2}}{y}$ on the upper half plane $\mathbb{H}$. Pushing forward the hyperbolic metric under $f$ will give you the Riemannian metric you desire.
The index 6 subgroup is generated by the two parabolic transformation $$z \mapsto z+2 $$ $$z \mapsto \frac{1}{\frac{1}{z} + 2} = \frac{z}{z+2} $$ It has a fundamental domain which is the ideal quadrilateral in $\mathbb{H}$ with ideal vertices $-1$, $0$, $+1$, $\infty$. That quadrilateral is a union of 6 translates of the standard fundamental domain for $\text{SL}(2,\mathbb{Z})$, namely the triangle with vertices $i$, $\frac{1}{2} + \frac{\sqrt{3}}{2} i$, $\infty$ (which is one way to see that the index equals $6$).
The universal covering map $f$, when restricted to the ideal triangle with infinite vertices $0,1,\infty$, maps the interior of that triangle to the upper half plane fixing the three points $0,1,\infty$. I think that information is enough to let one construct an explicit formula for $f(z)$. It'll have some complex logarithms in it, in order to convert the angle $0$ corners into angle $\pi$ corners at each of $0,1,\infty$. There might be a complex analyst around here who knows the explicit formula for $f(z)$, or who knows a reference for it; unfortunately I don't know either of those.