I am having trouble applying Ito's Formula to the following:
Let $Z_t = W_{1t}^2 e^{W_{1t}+ \int_0^t W_{3s}dW_{2s}}$. Find $dZ_t$. $W_1,W_2,W_3$ are independent Brownian motions.
I know the formula but I am having trouble differentiating the integral with respect to $W_2$ and $W_3$.
Let $A_t := \int_0^t W_{3,s} dW_{2,s}$. As pointed out by @fesman in the comments, we have $dA_t = W_{3,t}dW_{2,t}$. Note that this is just notation for how we defined $A_t$, we are not actually differentiating anything. Then we have $Z_t = W_{1,t}^2 e^{W_{1,t} + A_t}$, and Ito's formula gives
\begin{align*} dZ_t &= (W_{1,t}^2 e^{W_{1,t}+A_t} + 2 W_{1,t} e^{W_{1,t}+A_t}) dW_{1,t} + W_{1,t}^2 e^{W_{1,t}+A_t} dA_t \\ & \qquad + (W_{1,t}^2 e^{W_{1,t}+A_t} + 4 W_{1,t} e^{W_{1,t}+A_t} +2e^{W_{1,t}+A_t}) dW_{1,t}dW_{1,t} + W_{1,t}^2 e^{W_{1,t}+A_t} dA_t dA_t \\ &= (W_{1,t}^2 e^{W_{1,t}+A_t} + 2 W_{1,t} e^{W_{1,t}+A_t}) dW_{1,t} + W_{1,t}^2 e^{W_{1,t}+A_t} W_{3,t}dW_{2,t} \\ & \qquad+ (W_{1,t}^2 e^{W_{1,t}+A_t} + 4 W_{1,t} e^{W_{1,t}+A_t} +2e^{W_{1,t}+A_t}) dt + W_{1,t}^2 e^{W_{1,t}+A_t} W_{3,t}^2dt. \end{align*}
Again, to emphasize, we have not differentiated $Z_t$. This is just notation for $$Z_T = \int_0^T (W_{1,t}^2 e^{W_{1,t}+A_t} + 2 W_{1,t} e^{W_{1,t}+A_t}) dW_{1,t} +\int_0^T W_{1,t}^2 e^{W_{1,t}+A_t} W_{3,t}dW_{2,t} + \int_0^T (W_{1,t}^2 e^{W_{1,t}+A_t} + 4 W_{1,t} e^{W_{1,t}+A_t} +2e^{W_{1,t}+A_t} + W_{1,t}^2 e^{W_{1,t}+A_t} W_{3,t}^2) dt.$$