if $\mathbb{R}^+$ is the set of non-negative real numbers, your condition implies that $$(f(x)-f(0)) (f(\frac{x}2)-f(0)) = 0$$ for all positive $x,$ which is pretty convincing (setting $g(x) = f(x) - f(0),$ the function $g$ will be identically $0.$).

if $\mathbb{R}^+$ is the set of non-negative real numbers, your condition implies that $$(f(x)-f(0)) (f(\frac{x}2)-f(0)) = 0$$ for all positive $x,$ which is pretty convincing (setting $g(x) = f(x) - f(0),$ the function $g$ will be identically $0.$).

Addendum Since the map $(x, y) -> (x+y)/2, \sqrt{x y})$ is surjective onto the set $x>y,$ if your function were real analytic and nonconstant, then the set where your condition held would be a subvariety of $R^+ \times R^+,$ so assuming more regularity makes the problem easy, as pointed out in comments.

if $\mathbb{R}^+$ is the set of non-negative real numbers, your condition implies that $$f(x) f(\frac{x}2) = 0$$ for all positive $x,$ which is pretty convincing.

if $\mathbb{R}^+$ is the set of non-negative real numbers, your condition implies that $$(f(x)-f(0)) (f(\frac{x}2)-f(0)) = 0$$ for all positive $x,$ which is pretty convincing (setting $g(x) = f(x) - f(0),$ the function $g$ will be identically $0.$).