Consider a geometric Brownian motion described by the SDE:
$$dS_t = \mu S_t dt + \sigma S_t dB_t$$
where $B_t$ is a Brownian motion and $\mu$ and $\sigma$ are constants. We have that:
$$d(\ln S_t) = \sigma dB_t +\left(\mu - \frac{\sigma^2}{2}\right)dt$$
from which we immediately obtain:
$$S_t = S_0 \exp\left[(\sigma B_t)+\left(\mu - \frac{\sigma^2}{2}\right)t\right]$$
Consider now the process $R_t = S_t^{-1}$; in this case, since $dR_t = - S_t^{-2} dS_t$, we obtain the SDE:
$$dR_t = -\mu R_t dt - \sigma R_t dB_t$$
and by applying the same procedure, we obtain:
$$R_t = R_0 \exp\left[(- \sigma B_t)+\left(- \mu - \frac{\sigma^2}{2}\right)t\right]$$
On the other hand, from inverting immediately the $S_t$ solution I would obtain:
$$R_t = R_0 \exp\left[(- \sigma B_t)+\left(- \mu + \frac{\sigma^2}{2}\right)t\right]$$
Obviously there's something wrong here, but which method is correct? And where's the mistake?