Given a process for three independent Brownian motions $B_{1t}, B_{2t}, B_{3t}$
$$Y_t = B^2_{1t}e^{B_{1t} + \int_0^t B_{3s}dB_{2s}}$$
We need to find $dY_t$
I start by applying Ito's lemma:
$$dY_t = 2B_{1t}e^{B_{1t} + \int_0^t B_{3s}dB_{2s}} dB_{1t} + B_{1t}^2 e^{B_{1t} + \int_0^t B_{3s}dB_{2s}} d B_{1t} + …$$
Instead of dots, there should be derivatives wrt to $dB_{2t}$ and $dB_{3t}$. However, given that they are in integral, I am not really sure how to proceed further.
Edit:
Let $e^{B_{1t} + \int_0^t B_{3s}dB_{2s}}$ be $\tilde e$.
After applying the hint $dZ = B_3dB_2$ I get:
$$dY_t = 2B_1 \tilde e dB_1 + B_1^2 \tilde e dB_1 + B_1^2 \tilde e dZ + \frac 12 (2 \tilde e + 2 B_1\tilde e + B_1^2 \tilde e) dt =$$
$$
\frac{\mathrm{d}Y_t}{Y_t} = \left(\frac{2}{B_{1}}+1\right)\mathrm{d}B_{1} + B_{3}\mathrm{d}B_{2} + (\frac {1}{B_1^2} + \frac{1}{B_1} + \frac 12)dt
$$
Best Answer
You have simply : $$ \mathrm{d}\left(\int_0^tB_{3,s}\mathrm{d}B_{2,s}\right) = B_{3,t}\mathrm{d}B_{2,t} $$ Indeed, you can convince yourself by integrating back this expression. Otherwise, you can see it as Leibniz's rule for differentiation under the integral sign for the parameters $t,B_{1,t},B_{2,t},B_{3,t}$ simultaneously $-$ but only $t$ appears in the expression above, since $s,B_{1,s},B_{2,s},B_{3,s}$ are dummy variables.
In consequence, the final result is given by $$ \frac{\mathrm{d}Y_t}{Y_t} = \left(\frac{1}{B_{1,t}^2}+\frac{2}{B_{1,t}}+\frac{1}{2}\right)\mathrm{d}t + \left(\frac{2}{B_{1,t}}+1\right)\mathrm{d}B_{1,t} + B_{3,t}\mathrm{d}B_{2,t} $$