I have to show that $$ Y_t = e^{\theta N_t - \frac{1}{2}\theta^2 \int^t_0 g(s)^2 ds} $$ where $ N_t=\int^t_0g(s)dW_s $ satisfies $$ dX_t = \theta f(t) X_t dW_t $$ and that $ N_t \sim N(0, \int^t_0 g(s)^2 ds) $.
I have been trying to follow along with the process given here; however, I am uncertain as to how to go about without having the $dt$ term and the intergral within the $Y_t$ exponential makes me a bit confused.
Let $f(x,y) = e^{\theta x - \frac12 \theta^2 y}$ so that $Y_t = f(N_t, \int_0^t g(s)^2 ds)$. Let's also notice here that $\int_0^t g(s)^2 ds = \langle N \rangle_t$. By Ito's lemma, applied to $f$ we find \begin{align} dY_t =& \frac{\partial f}{\partial x}(N_t, \langle N \rangle_t) dN_t + \frac{\partial f}{\partial y}(N_t, \langle N \rangle_t) d \langle N \rangle_t + \frac12 \frac{\partial^2 f}{\partial x^2}(N_t, \langle N \rangle_t)d \langle N \rangle_t \\ =& \theta Y_t dN_t - \frac12 \theta^2 Y_t d\langle N \rangle_t + \frac12 \theta^2 Y_t d \langle N \rangle_t \\=& \theta Y_t dN_t \\=& \theta Y_t g(t) dW_t \end{align} where we don't see any other terms in the first line since $\langle N, \langle N \rangle \rangle = \langle \langle N \rangle, \langle N \rangle \rangle = 0$ and the last line follows by associativity of the integral since $dN_t = g(t) dW_t$.