If we are given that $$ Z_{t}=\sigma N_{t}dW_{t}+\sigma W_{t}dN_{t} $$ where $W_{t}$ is a Brownian motion, $N_{t}$ is a Poisson process with intensity $\lambda$, and both are independent.
How do apply Ito lemma on this? So far I've got: $$ df(t,Z_t)=\left[\frac{\partial f}{\partial t}+\frac{1}{2}\sigma^{2}N_{t}^{2}\frac{\partial^{2}f}{\partial z^{2}}\right]dt+\left[\sigma N_{t}\frac{\partial f}{\partial z}\right]dW_{t}+\left[f\left(t,Z_{t^{-}}+\sigma W_{t}\right)-f\left(t,Z_{t^{-}}\right)\right]dN_{t} $$ Is this correct? I am not sure about the coefficient on $\frac{\partial^{2}f}{\partial z^{2}}$. The result I am seeking to obtain is just $\frac{1}{2}\sigma^{2}\frac{\partial^{2}f}{\partial z^{2}}$, and not $\frac{1}{2}\sigma^{2}N_{t}^{2}\frac{\partial^{2}f}{\partial z^{2}}$.
Thanks.
Most likely you meant $dZ_t=\sigma N_t\,dW_t+\sigma W_t\,dN_t\,.$ For non continuous processes Ito's lemma is better written in this form \begin{align} &f(T,Z_T)=f(0,Z_0)+\int_0^T\left[\frac{\partial f}{\partial t}+\frac{1}{2}\sigma^{2}N_{t}^{2}\frac{\partial^{2}f}{\partial z^{2}}\right](t,Z_t)\,dt\\&+\int_0^T\sigma N_{t}\frac{\partial f}{\partial z}(t,Z_{t-})\,dW_t\\ \tag{1} &+\sum_{0\le t\le T}\Big[f(t,Z_t)-f(t,Z_{t-})\Big]-\sum_{0\le t\le T}\frac{\partial f}{\partial z}(t,Z_{t-})(Z_t-Z_{t-})\,. \end{align} Your coefficient of $\frac{\partial^{2}f}{\partial z^{2}}$ is correct.
Note however that $Z_t=\sigma N_t W_t\,.$ Writing $\Delta N_t=N_t-N_{t-}$ we get therefore \begin{align} Z_t-Z_{t-}&=\sigma W_t\Delta N_t\,,\\ f(t,Z_t)-f(t,Z_{t-})&=\Big[f(t,Z_t)-f(t,Z_{t-})\Big]\Delta N_t\\ &=\Big[f(t,Z_{t-}+\sigma W_t)-f(t,Z_{t-})\Big]\Delta N_t \end{align} because $\Delta N_t$ is either zero or one. Recall also that $$ \sum_{0\le t\le T}X_{t-}\,\Delta N_t=\int_0^TX_{t-}\,dN_t\,. $$ This allows to write the last two terms in (1) as \begin{align}\tag{2} \int_0^T\Big[f(t,Z_{t-}+\sigma W_t)-f(t,Z_{t-})\Big]\,dN_t-\int_0^T\frac{\partial f}{\partial z}(t,Z_{t-})\,\sigma W_t\,dN_t\,. \end{align} In your equation the last term of (2) was missing.