Sources: Calculus: Early Transcendentals (6 edn 2007). p. 890, Section 14.3. Exercise 50b, c.
This user wrote that $z_1 = f(xy)$ and $z_2 = f(x/y)$ depend
on only one variable - there's no comma between the parentheses.
Defining $t_1 = xy$ transforms $f(xy)$ into $f(t_1)$, and $t_2 = x/y$ transforms $f(x/y)$ into $f(t_2)$. Yes, $f(t_1)$ and $f(t_2)$ are single-variable.
But doesn't this misrepresent the original 2 independent variables ($x, y$)? These definitions don't change the original truth that $f$ did depend on 2 independent variables. So $f$ is multivariable.
The function $f: X \mapsto f(X)$ is not multivariable. The function $m: (X,Y) \mapsto XY $ ("multiplication", if you like) is multivariable. The function $g:(X,Y) \mapsto Z_1$ is multivariable, and given by $g = f \circ m$, so $$g(x,y)=f(m(x,y))=f(xy). $$