= Solution
For fixed endpoints, the <Euler-Lagrange equation> is
$$
\boxed{\frac{d}{dx}f_{y'}-f_y=0.}
$$
Let the sphere have radius $R$ and write the speed as $v_0\sin\theta$, with $v_0>0$. The infinitesimal travel time in the interior $0<\theta<\pi$ is
$$
dt=\frac R{v_0}\sqrt{\frac{d\theta^2}{\sin^2\theta}+d\phi^2}.
$$
This is the <flat-cylinder travel metric for sine-latitude speed>. Introduce
$$
z=\operatorname{arsinh}(\cot\theta),\qquad dz=-\frac{d\theta}{\sin\theta}.
$$
The travel-time metric becomes $(R/v_0)\sqrt{dz^2+d\phi^2}$, an ordinary Euclidean metric on a cylinder because $\phi$ is periodic. After choosing a lift of $\phi$, shortest paths are straight segments in $(\phi,z)$. For a route that is a graph over $\phi$, the <Euler-Lagrange equation> for $\sqrt{1+(dz/d\phi)^2}$ reads
$$
\frac{d}{d\phi}\left(\frac{z'}{\sqrt{1+z'^2}}\right)=0.
$$
Thus $z'=A$ is constant and $z=A\phi+B$. Transforming back,
$$
\boxed{\sinh(A\phi+B)=\cot\theta.}
$$
This proves the requested form for nonmeridional routes, including the equator when $A=B=0$. It is also globally minimizing for its chosen azimuthal lift: Euclidean arc length is at least the straight-line separation, and the shortest lift minimizes that separation among angular differences differing by $2\pi$.
There is a coordinate exception to the printed universal form. When the shortest angular separation is zero, the fastest path is the meridian $\phi=\text{constant}$; it is a vertical straight line on the cylinder and cannot be represented by finite $A,B$ as a graph over $\phi$. The poles correspond to $z=\pm\infty$ and cannot be reached in finite time with this speed law.
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