I want to show that the Itô-process
$$ Z_t = \text{exp}( \int_0^t \theta_s dB_s - \int_0^t \frac{1}{2} \theta_s^2 ds) $$
is a solution to the SDE
$$ dX_t = \theta(t)X_t dB_t$$
using Itô's formula,
$$ dg(t,B_t) = \frac{\partial g}{\partial x}(t,B_t)dB_t + ( \frac{\partial g}{\partial t} (t,B_t) + \frac{1}{2} \frac{\partial^2 g}{\partial x^2} (t,B_t))dt $$
i.e, I want to find a function $g(t,x)$ s.t $ \frac{\partial g}{\partial x}(t,B_t) = \theta(t) g(t,B_t)$ and $ \frac{\partial g}{\partial t} (t,B_t) + \frac{1}{2} \frac{\partial^2 g}{\partial x^2} (t,B_t) = 0$.
But I'm stuck. How do I proceed?
Consider the following Itô process given by $$dY_t = - \frac{\theta_t^2}{2} dt + \theta_t dB_t $$ Now, to derive the dynamics of $Z_t$ you can apply Itô's formula to the function $f(x) = e^x$. Indeed, $$dZ_t = df(Y_t) = \Big\{- \frac{1}{2} \theta_t^2e^{Y_t} + \frac{1}{2}\theta_t^2 e^{Y_t} \Big\} dt + \theta_t e^{Y_t} dB_t = \theta_t e^{Y_t} dB_t = \theta_t Z_t dB_t$$ So, indeed $Z_t$ satisfies your SDE.