 Solution

Away from the three poles, the identity from the preceding part puts $F(u)$ on the cubic $Q=0$. The local extensions and continuity put the three pole images there too. Thus $F(C)\subseteq \{Q=0\}$, a cubic curve.

The local calculations gaveThey are three distinct points on the line with first coordinate zero. Hence $F(p),F(q),F(r)$ are collinear. This is three-point Riemann-Roch construction of a plane cubic.