= Solution
The <Strong Markov property> and part (b) show that, before $\tau=\tau_x\wedge\tau_y$,
$$
F(V_t)=\mathbb P(\tau_x>\tau_y\mid\mathcal F_t).
$$
Thus $F(V_{t\wedge\tau})$ is a bounded <martingale>. From the given stochastic differential equation,
$$
d[V]_t=\frac{(1-V_t)^2}{Y_t^2}dt.
$$
The <Itô formula> says that the drift of $F(V_t)$ is
$$
\frac1{2Y_t^2}\left[(1-v)^2F''(v)+\left(\frac{d-1}{v}+(3-d)v-2\right)F'(v)\right]_{v=V_t}dt.
$$
It must vanish. Dividing by $(1-v)^2/2$ and using the algebraic identity supplied in the question gives
$$
F''(v)+\left(\frac{d-1}{v}+\frac{2d-4}{1-v}\right)F'(v)=0,
\qquad v<0.
$$
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