Is there a better, more rigorous explanation for why this partial derivative is 0 using Ito's Lemma?

I encountered the following slide in a lecture on Ito's Lemma.

The lecturer explained that $$\frac{\partial V}{\partial t} = 0$$ because the first two derivatives on the slide already took into account time into the change of the value of V.

I'm not convinced. If $V = \log S(t)$ is a function of time, why wouldn't we have to use the chain rule for the third derivative on the slide?

$$\frac{\partial V}{\partial t} = \frac{\partial V}{\partial S(t)} \cdot \frac{\partial S(t)}{\partial t} = S^{-1} \cdot \frac{\partial S(t)}{\partial t} = ...$$

I'm not sure where to go from here to show that it is in fact 0.

Answer

A process indeed depends on time $t$. However, in Ito's lemma, only derivatives with respect to independent time variable $t$ is considered. That is, for a process of the form $f(S_t, t)$, $\frac{\partial f}{\partial t}$ is the derivative with respect to the second, that is, the independent, $t$ variable, however, the parameter $t$ in the process $S_t$ is not considered. Ito's lemma takes a particular form, which can not be understood in the normal calculus sense.