Consider two three stochastic processes $X$, $Y$ and $Z$ in probability space $(\Omega, (\mathcal F_t)_{t \geq0},\mathbb P)$ such that
$$ X_t = \exp\left(\int_0^t f_s ds\right), $$
$$ Y_t = \exp\left(\int_0^t g_s ds\right), $$ and $$ Z_t = Z_0 \exp\left(\int_0^t \left(g_s - f_s - \frac{1}{2} \sigma^2 \right)ds + \int _0^t\sigma_s dW_s\right) $$ where $f$ and $g$ are stochastic process as well such that $X$ and $Y$ are not independent and $\sigma$ is deterministic (Let's assume all desired integrability conditions apply for all process. Also W is a $\mathbb P$-Brownian motion and $(\mathcal F_t)$ is its natural filtration).
Now consider an equivalent measure $\mathbb Q \sim \mathbb P$ whose Radon-Nikodyn derivative is $$ \left.\frac{d\mathbb Q}{d\mathbb P}\right|_{\mathcal F_t} = \mathcal E_t(\int_0^{\cdot}\sigma_s dW_s)=\frac{X_tZ_t}{Y_t Z_0}. $$
I would like to show that $$ \mathbb E^{\mathbb P}\left[ Z_T | \mathcal F_t \right] = Z_t \frac{X_t}{Y_t}\frac{ \mathbb E^{\mathbb P}\left[ X_T^{-1} | \mathcal F_t\right] }{\mathbb E^{\mathbb P}\left[ Y_T^{-1}| \mathcal F_t\right]}. $$ To do so I would like to do (even if it is not not necessary; just as an exercise) by passing by the measure $\mathbb Q$. At some point I end up with the following $$ \mathbb E^{\mathbb P}\left[ Z_T | \mathcal F_t \right] = Z_t \frac{X_t}{Y_t} \mathbb E^{\mathbb Q}\left[ \frac{Y_T}{X_T} | \mathcal F_t\right]. $$ and I am stuck.
Could someone help from there?