Prove that if $h\neq k\in\mathbb{R}$ and $f$ is a real-valued function so that $f(x)f(y) = y^h f(x/2) + x^k f(y/2)$ for all positive reals $x,y$, then $f$ must be identically zero.
If one equates $f(y)f(x) = f(x)f(y)$ (using the original functional equation), one gets that for $x\neq 1, f(x/2) = A((2x)^h - (2x)^k)$ for some constant $A$. If one assumes that the coefficients of terms with equal exponents must match, then we get $A=0$ (in more detail, we equate $A^2((2x)^h - (2x)^k)((2y)^h - (2x)^k) = (x^h - x^k) A y^h + x^k A (y^h - y^k).$ But since $h$ and $k$ are real, it's not clear that equating coefficients on both sides is even valid. For polynomials, of course it would be valid since nonconstant polynomials only have finitely many roots.
I'm not sure if the set $\{x^i : i \in \mathbb{R}\}$ linearly independent over $\mathbb{R}$. If the given set is in fact linearly independent, then for any $k\ge 1$ and constants $c_1,\cdots, c_k\in \mathbb{R}$, if $c_1 x^{i_1} + c_2 x^{i_2}+\cdots + c_k x^{i_k} = 0$ (i.e. it's the zero function) then $c_1=c_2=\cdots = c_k = 0.$
Partial Solution:
First, note that $$f(0)^2 = 0^hf(0) + 0^kf(0) = 0.$$
The remainder of this answer assumes that $f(1) = 0$. Now with $x = y = 1$, $$0 = f(1/2)+f(1/2) \implies f(1/2) = 0.$$
Finally, for arbitary $x$ (set $y = 1$), $$0 = f(2x)f(1) = f(x)+(2x)^kf(1/2) = f(x).$$ Therefore, $$f(x) =0 \quad \forall x \in \mathbb{R}^+.$$