$$a>0,$$ $$b>0,$$ $$\sigma >0$$ $X$ is the solution of : $$dX_t=aX_t(b-X_t)\,dt+\sigma X_t \, dB_t,\quad X_{0}=1 $$ I have also shown before that $$L_t=e^{(ab-\sigma^2/2)t+\sigma B_t}$$
Now we have $$Y_t=L_t/X_t$$ and I have to calculate $$dY_{t} $$ Is it correct to write that : $$dY_t =(dL_t X_t-dX_t L_t)/X_t^2\text{ ??} $$
Thanks in advance for your help
Since $L_t$ is a.s. positive and the function $f(x) = 1/x$ is $C^2$ on $(0,\infty)$ you can use Ito as you want to. You have messed up the algebra though. There should definitely be an $L^2$ on the bottom, not an $X^2$.
With your edited post you need to be more careful. You need to show that $X_t$ is a.s. positive if you want to apply Ito like you usually would.