Let $x, y \in (-\infty, \infty)$
I know that $f(x,y) = xy$ has a equilibrium at $(0,0)$, but I do not know how to logically arrive at this solution.
I know that a Nash equilibrium is when $\min_x\max_y xy = \max_y \min_x xy$
Start from the LHS, we have, $$\min_x\max_y xy$$ The inner player goes first, which implies $y$ player will choose $\infty$, and $x$ will choose $-\infty$, yielding a value of $-\infty$.
For, $$\max_y \min_x xy$$
The inner player goes first and chooses $x = -\infty$, and outer player y will choose $-\infty$, yielding $\infty$.
It does not seem to me that the players can arrive at $(0,0)$. Because whoever goes second sets the value of the game.
What is missing?
If $x > 0$, then $xy$ can be made as large and positive as we want, if we choose $y$ to be large and positive. So, for $x > 0$, $\sup_y xy = \infty$.
If $x < 0$, then $xy$ can still be made as large and positive as we want, this time choosing $y$ to be large and negative. So, for $x < 0$, $\sup_y xy = \infty$.
If $x = 0$, then no matter what $y$ chooses, the result of $xy$ is $0$. Thus, if $x = 0$, $\sup_y xy = 0$. Therefore, $$\sup_y xy = \begin{cases} 0 & \text{if } x = 0 \\ \infty & \text{otherwise.}\end{cases}$$
Now, if we are taking the infimum (or indeed minimum) of this extended real function over all possible choices for $x$, we clearly need to choose $x = 0$, for a minimum of $0$.
Similar logic shows that $$\inf_x xy = \begin{cases} 0 & \text{if } y = 0 \\ -\infty & \text{otherwise.}\end{cases}$$ Thus, taking the maximum over $y$, we get $0$, which occurs when $x = 0$. Therefore, $(0, 0)$ is the Nash equilibrium.