Let $S$ be a stochastic process
$dS=\mu S dt + \sigma S dW \\ S(0)=s_0$
and let $u=u(S(t),t)$ be some regular enough function and let $\phi(t)=u_s(S(t),t)$ be the derivative of $u$ w.r.t. s.
I cannot find why we have:
$d\phi \ dS=\sigma^2S^2u_{ss}dt$
The assertion is in L.C. Evan's course: "An introduction to SDE".
Here is a simple way to see it. Let $S$ be geometric Brownian motion (so that $dS=\mu S\,dt+\sigma S\,dB_{t}$) and define $\phi=u_s(S,t)=\partial_s u(S,t)$.
Use Ito's Lemma: $$ d\phi=\underbrace{\left( u_{st} +\mu S u_{ss} + \frac{\sigma^2S^2}{2}u_{sss} \right)}_{\displaystyle\Psi}dt + \sigma S u_{ss}\,dB_t $$
Then use the rules for stochastic differentials (see here and here) to get: \begin{align} dS\,d\phi &= (\Psi dt + \sigma u_{ss}\,dB_t)(\mu S dt + \sigma S\,dB_t)\\[1.7mm] &= \Psi\mu\underbrace{dt\,dt}_0 \;+\, \sigma\mu\underbrace{dB_t\,dt}_0 \;+\, \Psi\sigma S\underbrace{dt\,dB_t}_0\,+\, u_{ss}\sigma S\sigma S \underbrace{(dB_t)^2}_{dt}\\ &= \sigma^2S^2 u_{ss} \,dt \end{align}
