Consider $Y_t=B_t^3$, $t \geq 0$ where $(B_t)_{t \geq 0}$ is standard Brownian Motion.
Here, $dB_t = 0+1\cdot dB_t$ and $G(x,t)=x^3$, What is $dY_t$?
Using Ito's Lemma, we can calculate the partial derivatives and get:
$$\begin{align*} dY_t &= [3B_t^3\cdot 0+0\cdot \frac{1}{2}\cdot 6B_t\cdot 1^2]dt + 3B_t^2\cdot 1\cdot dB_t \\ &= 3B_t^2dt + 3B_t^2dB_t \end{align*}$$
If $Y_t = t\cdot B_t$, how can I use this same logic to get $d(tB_t)$?