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 $$Z_T = \int_0^T (W_{1,t}^2 e^{W_{1,t}+A_t} + 2 W_{1,t} e^{W_{1,t}+A_t}) dW_{1,t} +\int_0^T W_{1,t}^2 e^{W_{1,t}+A_t} W_{3,t}dW_{2,t} + \int_0^T (W_{1,t}^2 e^{W_{1,t}+A_t} + 4 W_{1,t} e^{W_{1,t}+A_t} +2e^{W_{1,t}+A_t} + W_{1,t}^2 e^{W_{1,t}+A_t} W_{3,t}^2) dt.$$