= Solution
The angle $\theta$ between the oriented coordinate tangent vectors satisfies $\cos\theta=F/\sqrt{EG}$. The constant-angle assumption therefore gives $F=k\sqrt E\sqrt G$ for a constant $k$. The <first fundamental form> is positive definite, so $EG-F^2>0$ and hence $|k|<1$. Substitution into the <geodesic coordinate grid> identities from part (i) gives
$$
k(\sqrt E)_v=(\sqrt G)_u,\qquad k(\sqrt G)_u=(\sqrt E)_v.
$$
Eliminating either derivative yields $(1-k^2)(\sqrt E)_v=0$ and $(1-k^2)(\sqrt G)_u=0$. Since $E,G>0$ and $1-k^2>0$, this proves
$$
\boxed{E_v=0=G_u.}
$$
If the angle is understood between unoriented curves, its constancy initially fixes $|F|/\sqrt{EG}$; continuity fixes the sign on each connected chart when this quantity is nonzero, and the zero case already has $F=0$. Thus the same conclusion holds. In fact <constant-angle geodesic coordinate grids are locally flat>, although only the displayed derivatives are needed here.
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