I don't think this (nonlinear) stability problem is simple because the SDE for $Y_t$ has multiplicative noise and time-dependent coefficients. However, the result seems plausible for the following reason: when $Y_0$ is close to zero, $Y_t$ is well approximated in a strong sense by the process $\tilde Y_t$ which satisfies the linear SDE: $$ d \tilde Y_t = X_t \tilde Y_t dt + X_t \tilde Y_t d V_t \;. $$ This linear SDE has the pathwise solution: $$ \tilde Y_t = Y_0 e^{ \int_0^t X_s dV_s + \int_0^t ( X_s - \frac{1}{2} X_s^2) ds } \;. $$ Since almost surely $$ \frac{1}{t} \left( \int_0^t X_s dV_s + \int_0^t ( X_s - \frac{1}{2} X_s^2) ds\right) \to - \frac{1}{4} \quad \text{as $t \to \infty$ } $$ it follows that almost surely $\tilde Y_t \to 0$ as $t \to \infty$. A similar argument holds if $Y_0$ is close to one. ---------- I think filling in the details of this argument will require more work. First, for $Y_0$ away from zero or one, one would have to argue that $Y_t$ almost surely reaches any neighborhood of zero or one, where the linear argument given above is valid. (This seems a bit tricky because of the complicated time-dependence of the coefficients.) Second, one needs a comparison argument to prove that $\tilde Y_t \to 0$ almost surely implies that $Y_t \to 0$ with high probability. (A problem can happen if $\tilde Y_t$ itself leaves a neighborhood of zero in which case the linear approximation is no longer valid.)