> f(x, y) = 0. With that you can plot the graph of f(x, y) either as a 3D surface with f(x, y) being the height at point (x, y)

If f(x, y) = 0, wouldn’t using f(x, y) for the height just result in a flat graph?

f(x, y) = 0 is true only for some combinations of x and y. It’s an equation to be solved, not a universal statement like ∀ x, y : f(x, y) = 0, nor a definition like f(x, y) ≔ 0 (or “≝”). The solutions to the equation are the points (x, y) where the graph has height 0. Which points these are depends on how f is defined.

For example, f might be defined as f(x, y) ≔ x² + y² – 1. Then the points (x, y) for which f(x, y) = 0 are those on the unit circle (those for which x² + y² = 1). The graph will have height 0 only for those points.

They're really two different types of equal signs.

f(x,y) = x+y might be better written as f(x,y) := x+y where := means "is defined as". Then f(x,y) = 0 is an equation that expands to x+y = 0, or in familiar intro algebra form, y=-x.

g(x,y) := 0 really is a flat plane.

When we say "f(x, y) = 0" in this context, we also usually have a separate definition for f(x, y) provided, where that f(x, y) is not necessarily 0 at for all x,y. And so this constraint "f(x, y) = 0" means "find pairs of x and y such that it makes f(x, y) become 0".

If "f(x, y) = 0" is actually the definition of f(x, y), then yes, it would be a pretty boring graph.