For a stochastic process $X(t)$ and Brownian motion $Z(t)$
$$ d X = a(X, t) d t + b(X,t) d Z $$
where $d Z \sim \mathcal{N}(0, d t)$. Ito's lemma is defined as:
$$ d f = \left( \frac{\partial f}{\partial t} + a\frac{\partial f}{\partial X} + \frac{b^2}{2} \frac{\partial^2 f}{\partial X^2} \right) d t + b \frac{\partial f}{\partial X} d Z $$ where $a$ and $b$ are shorthand for $a(X,t)$ and $b(X,t)$.
If we apply this to a process $X(t)$ defined as $d X = \mu X d t + \sigma X d Z$ with $f(X,t) = \ln(X)$ then the book I am reading states that $\frac{\partial f}{\partial t} = 0$. I do not understand why the partial derivative of $f()$ wrt $t$ is 0 when f is a function of $X(t)$ which is a function of $t$.
In particular, $\frac{\partial f}{\partial t} = \frac{\partial f}{\partial X} \frac{d X}{d t}$ and neither of them are 0 so how can $\frac{\partial f}{\partial t} $ be 0? I can only assume that this chain rule application is wrong and this must be because the chain rule does not work with random variables?
I think you're overthinking this: $\frac{\partial f}{\partial t}$ is a partial derivative and so you hold to the $x$ coordinate constant. More specifically, it's perhaps more clear if its written in a more formal fashion.
Let $f:\mathbb{R}^2 \to \mathbb{R}$ and let $f_t,f_x$ and $f_{xx}$ denote its partial derivatives. Then Ito's formula says:
$$d f(t,X_t) = \left(f_t(t,X_t) + a f_x(t,X_t) + \frac{b^2}{2} f_{xx}(t,X_t) \right)\,dt + b f_x(t,X_t)\,dZ\,.$$
In other words, the point of Ito's formula is that those derivatives are just considering $f$ as a function.