But why did you use Exp[-x^2] for f[x]? Where do you get it from?

ContourPlot needs a function to evaluate; try

ContourPlot[unknownFunction[a, b], {a, -1, 1}, {b, -1/2, 1/2}]

to see nothing. Same thing with

Clear[f, g]
(* f[x_]:= Exp[-x^2] *)
g[x_, y_] := f[3 x - y]
ContourPlot[g[a, b], {a, -1, 1}, {b, -1/2, 1/2}]

again you see nothing because g does not evaluate to some real number. By the way, you may type directly with your function f (let's say f = BesselJ for the sake of definiteness)

ContourPlot[BesselJ[1, 3 a  - b], {a, -1, 1}, {b, -1/2, 1/2}]

CountourPlot does not need a function of two variables, but it needs 2 parameters (a and b in the last example). You can do also

ContourPlot[BesselJ[1, 3 a], {a, -1, 1}, {b, -1/2, 1/2}]

now the contour plot is real 1-D and even

ContourPlot[BesselJ[1, \[Pi]], {a, -1, 1}, {b, -1/2, 1/2}]

can be given: now the contour plot is contour-less, it is 0-D, the whole area evaluates to the same value BesselJ[1, Pi].

You say that f is a function of one variable but it consists of x and y?

One should distinguish between dependent and independent variables, to make it clear.

Is Sin[x0 + 2 x1 + 3 x2 + 4 x3 + 5 x4 + 6 x5] a function of one independent variable? Yes. But this single independent variable variable x = p[x0,x1,x2,x3,x4,x5] = x0 + 2 x1 + 3 x2 + 4 x3 + 5 x4 + 6 x5 is created by the map p: R^6 -> R. Each x represents a whole five dimensional linear space made of the six dependent variables x0, x1, x2, x3, x4, and x5 because of the equation established by the map p.

In the end, if somebody talks about a function as a function of one variable, he or she most often means a function of one independent variable.

Another way to look into this is the Chain Rule invented by Leibniz, see also Chain Rule.

In[24]:= Clear[f, g]
f[x_] := Exp[-x^2]
g[x_, y_] := f[3 x - y]
ContourPlot[g[a, b], {a, -1, 1}, {b, -1/2, 1/2}]

does not work? Because f is a function of one variable, the contour plot is quasi-1D.