Note first that draws are not possible so any position $(x, y)$ is either a win or a loss for the player receiving it. Let $\phi$ be the positive root of $t^2 - t - 1 = 0$, so $\phi = \frac{1+\sqrt{5}}{2}$. Let $m = \min(x, y)$, $M = \max(x, y)$. We show that a player receiving $(x, y)$ wins if and only if $m = M$ or $M > \phi m$.

If $m \neq M$ and $M < \phi m$, then the player must pass on $(m, M-m) \neq (m, 0)$ and $m > \phi(M-m)$. So it is sufficient to show that a player receiving a position with $M > \phi m$ can either win or pass back a position with $\min m'$ and $\max M'$ such that $M' < \phi m'$ and $M' \neq m'$.

If $M > \phi m$, and $M$ is a multiple of $m$, then the player can pass on $(m, 0)$ and win. So assume $M > \phi m$ and $M = qm + r$ with $0 < r < m$. If $q \ge 2$, then the player can choose whether to pass on $(m, r)$ or $(m, m+r)$. If $(m, r)$ is a losing position, then he wins by passing that. If it is a winning position, then

he passes $(m, m+r)$ to the other player. The other player is now forced to pass back $(m, r)$, which is a winning position. So we may assume $q = 1$. Now the player passes $(m, r)$. We claim that $m < \phi r$ and $m \neq r$. Certainly $m \le r$. But $m+r > \phi m$, so $(\phi-1)m < r$, so $m < \phi r$.