Is there a difference between $f(x,y)$ $f(x;y)$ and $f(x\mid y)$? While reading I have come across all three of these notations seemingly at random, and as far as I can tell they are all positional arguments to a function, but I can't tell if they mean different things, do they?

$f(x,y)$ is just any function of the two variables; nothing else is implied.
$f(x|y)$ is usually used in probability, and is used in the context of "the probability that the first variable will be of some specified value $x$, given that the second variable was $y$. An example of this may make that clearer:
Say Charlie rolls 5 dice, three red and 2 blue.  And let's say tells us, in order, two numbers in this order:  the some of the two blue dice values, followed by the sum of the five dice values. Then we can define a joint probability distribution $f(x,y)$. For example,
$$f(11,29) = \frac{1}{18}\cdot \frac{1}{216}$$ because to get that, the blue die must at to $11$ and then all three red dice must be sixes.
Now Charlie is more of a tease; he only supplies the sum of the five dice values, $y$.  What is the probability of some number for $x$ given that the sum of the five values is $y$.  That is written as 
$$ f(x | y)$$
or more commonly, we say that we know that the value of $y$ is some $y_0$ and ask about
$$f(x | y = y_0)$$
For example, $f(12|y= 29) = \frac{3}{5}$ and $f(11, 29) = \frac{2}{5}$.
Finally, what will $f(x;y)$ mean?  In general, that sort of notation is used when talking about a one-parameter family of functions of one variable. For example, for $n$ an integer let $f(n)$ be the sum of the digits in $n$.  You can then imagine studying $f(n;b)$ where $b$ is the base in which you are working; the function we were talking about in the previous sentence would be $f(n;10)$.  You might feel this is a distinction without a difference but sometimes the notation makes things clearer. 
