We are given the following:
d$S_t$ = $\sin(S_t)t^2dt + e^{\sqrt{S_t}-t}dB_t$
And are asked to compute several different things, one of which is $d \log(S_t).$
If I'm understanding Ito's formula correctly, which I may not be, from the $dS_t$ given, we know:
(1) $f''(S_t) = 2t^2\sin(S_t)$, and
(2) $f'(S_t) = e^{\sqrt{S_t}-t}$
But then how do you determine $S_t$ to compute log($S_t$) and it's first and second derivatives?
$S_t$ is an Itō process, so you can apply the Itō-Doeblin formula to $f(t,x)=\log x.$ Therefore
$$d(\log(S_t))=\frac {1}{S_t}dS_t-\frac{1}{2S_t^2}dS_tdS_t$$
Note that $dB_tdB_t=dt,$ and $dtdB_t=0=dt^2, $ so $dS_tdS_t=e^{2(\sqrt{S_t}-t)}dt.$
Thus
$$d(\log(S_t))=(\frac{t^2\sin S_t}{S_t}-\frac{e^{2(\sqrt{S_t}-t)}}{2S_t^2})dt +\frac{e^{\sqrt{S_t}-t}}{S_t}dB_t.$$