 Solution

Take $\lambda =e^{\pi i/3}$ and the chosen $s=e^{\pi i/6}$. Then $\lambda s=i$, so $i/(\lambda s)=1$, and $-1/\lambda ={\lambda }^2$. In affine coordinates the chosen lift becomesSince ${\lambda }^2$ is a primitive cube root of unity, $1+{\lambda }^2+{\lambda }^4=0$. Three iterations fix $x$ and $y$, whereas one does not. ThusThe other square-root choice replaces this map by $(x,y)\mapsto (1+{\lambda }^{2}x,-y)$. Its cube is $(x,y)\mapsto (x,-y)$, so that lift has order six. This sign dependence is why a lift was specified explicitly.

Both fix the point at infinity. In the complex torus model their linear multipliers have order three or six, so the lattice is stable under a primitive cube root $\zeta =e^{2\pi i/3}$ and hence under $e^{\pi i/3}=1+\zeta$. Choose a shortest nonzero period $\omega$. The triangular lattice $\mathbb{Z}+\zeta \mathbb{Z}$ has covering radius $1/\sqrt{3}<1$: its equilateral triangles of side one each have circumradius $1/\sqrt{3}$. Approximating any period divided by $\omega$ by this lattice leaves a strictly shorter period, which must be zero. ThereforeThus the cubic has the hexagonal lattice shape, independently of which of the two lifts is chosen.