 Solution

The independent Gaussian increments of Brownian motion give $E[Y_t\mid {\mathcal{F}}_s]=Y_s{e}^{\mu (t-s)}$; the exponential is integrable at every finite time. ThereforeSimilarly $E[Y_t^{-1}\mid {\mathcal{F}}_s]=Y_s^{-1}{e}^{({\sigma }^2-\mu )(t-s)}$, so $\boxed{Y^{-1}\text{ is a martingale}⟺\mu ={\sigma }^2}$. This uses actual conditional expectations, not only the local stochastic differential equation.