This is the last part to a 3 part question! I am nearly done going through the questions I had difficulty with while studying, again, anyone's help would be greatly appreciated!
Let $X(t) = e^{r(T-t)}/S(t)$.
Find the SDE of $X(t)$ provided that $S(t)$ satisfies the BSM model.
Make sure the drift and diffusion terms are expressed in terms of $X(t)$ rather than $S(t)$.
Consider the function $f(x,t) = \frac {e^{r(T-t)}}{x}$. Then $X = f(t, S)$ and you can use Ito lemma to say that
$$dX_t = \frac{\partial}{\partial t}f(t,x)\mid_{ t,S} dt + \frac{\partial}{\partial x}f(t,x)\mid_{ t,S} dS_t+\frac 12\frac{\partial^2}{\partial x^2}f(t,x)\mid_{ t,S} d \langle S \rangle _t$$
Which results in
$$dX_t = \frac {-e^{r(T-t)}}{S_t}dt - \frac {e^{r(T-t)}}{S_t^2} dS_t + \frac {e^{r(T-t)}}{S_t^3} d \langle S \rangle _t$$
Given that $$\frac{dS_t}{S_t} = \mu dt + \sigma dW_t$$, we find
$$d \langle S \rangle_t = \sigma^2 S_t^2 dt$$ and substituting we find
$$dX_t = \frac {-e^{r(T-t)}}{S_t}dt - \frac {e^{r(T-t)}}{S_t} (\mu dt + \sigma dW_t) + \frac {e^{r(T-t)}}{S_t} \sigma^2 dt$$
Now given that $X_t = \frac {e^{r(T-t)}}{S_t}$ we find $$dX_t = -X_t dt - X_t(\mu dt + \sigma dW_t) + X_t \sigma^2 dt$$ or
$$\frac {dX_t}{X_t} = (\sigma^2 - \mu - 1)dt - \sigma dW_t$$