Now bear with me here, I'm not the best at math. I'm just trying to find out something that I never really learned. I was wondering, can a function have multiple inputs such as this one below?
$$f(x,y) = x \cdot y$$
[Math] Can functions have multiple inputs
functions
Related Solutions
What you are getting confused by most probably is the fact that a vector seems to be multiple elements, but really it's not, it's just a single element which (often) has multiple parts.
You can even have so called set-valued functions. An easy one to think of is the function which takes a positive integer to the set of its prime divisors. Let's call this function $\pi$.
The important thing in both these cases is that a given input only has one output. So in the case of a vector-valued function the output is the full vector (all of its parts). In the case of the set valued function for prime divisors you have to get the set of all the prime divisors. So for example $\pi(15)=\{3,5\}$ you can't have $\pi(15)=\{3\}$ or $\pi(15)=\{5\}$. There is only the one set containing all the divisors. It's the same situation with vector-valued functions. There is only one vector for any input. The function can't return just part of a vector. Thus we still have only one output for every input--it's just a bigger output.
As to why we don't define inverses as vector valued, that has a very easy reason. They wouldn't be inverses anymore. The definition of an inverse to a function $f$ is a function $g$ for which $g(f(x))=x$ if the value of $g$ is a vector containing $x$ that doesn't do what we want. It doesn't tell us the value of the original input to function $f$.
If $f$ is a function $A\to A$ where $A$ is a finite set, then $f$ is injective if and only it it is surjective. It is not possible for such an $f$ to be injective without being surjective or vice versa.
This can be proved by induction on the number of elements in $A$, but the details are slightly tedious. This property carries over to a function $f:A\to B$ where $A$ and $B$ have the same number of elements -- because "the same number of elements" means that there is a bijection $g:A\to B$, and then $f$ is injective/surjective exactly when $g^{-1}\circ f$ is, and $g^{-1}\circ f$ goes from $A$ to $A$ itself.
On the other hand, if $A$ is infinite, then it is possible for a function to be injective without being surjective, or vice versa. Some simple examples for $\mathbb N\to\mathbb N$ is $$ g(n) = n+7 \\ h(n) = \begin{cases} n & \text{if }n\le 42 \\ n-1 & \text{otherwise} \end{cases}$$ where $g$ is injective but not surjective, and $h$ is surjective but not injective.
(Strictly speaking this property of infinite sets is only guaranteed if the axiom of choice holds, but that is a techincal complication that you probably don't want to worry about at this level).
Best Answer
Short answer: Yes.
Long answer: Yes, but using the Cartesian product, you can consider multiple inputs as being a single input, where the single input is an ordered pair. So with your example, $f(x,y)=xy$, you can either consider it as a bivariate function with two inputs, $x$ and $y$; or as a univariate function with a single input, the pair $(x,y)$.