I asked earlier about applying Ito's lemma previous question, user @StratosFair gave me such answer, from which I didn't get one detail:"
Just notice that $Y_t$ is given by $Y_t \equiv e^{Z_t}$ where $Z_t$ is a solution of the SDE
$$dZ_t = h(t)dB_t - \frac12 h^2(t) dt .\tag1$$
Hence you can apply Itô's lemma to $f(t,Z_t) := e^{Z_t}$ as follows :
$$\begin{align*}dY_t = df(t,Z_t) &= f_t (t,Z_t) dt + f_x(t,Z_t)dZ_t + \frac{1}{2}h(t)^2f_{xx}(t,Z_t)dt\\ &= 0 + e^{Z_t}dZ_t + \frac{1}{2}h(t)^2 e^{Z_t}dt \end{align*}$$ "
Is it right to say, that derivative of f by t will be zero? If I have a look on example of applying that lemma, with geometrical brownian motion, example,I cant get, how does it lead to f_t = 0? Yes, in example with Geometrical BM we dont have t in f, but here we have relationship with t through h(t), isn't it so? I can't get it.
The function is $f(t,z)=e^z$, where $t,z$ are number arguments. The partial derivatives are for argument positions, not for the content of the arguments. Thus $f_t=0$, as there is no $t$ dependency on the right side.