Setting $y = 0$ gives
$$f^2(x)f(0) + f(0)^2f(2x) = 2f^2(x)f(0).$$
$$\text{Thus } f(2x) = \frac{f^2(x)}{f(0)}, \quad f(2y) = \frac{f^2(y)}{f(0)}. \qquad (1)$$
Substituting (1) into the given condition, we get
$$f(x)f(y)(f(x)f(y) - f(0)f(x+y)) = 0. \qquad (2)$$
From (2) and $f(x) \neq 0, f(y) \neq 0$, we get
$$f(x)f(y) - f(0)f(x+y) = 0 \qquad (3)$$
Let $g(x) = \frac{f(x)}{f(0)}$; then from (3) we know
$$g(x+y) = g(x)g(y). \qquad (4)$$
From (4) and mathematical induction, we obtain
$$g(nx) = (g(x))^n, \text{ for any positive integer } n. \qquad (5)$$
In (5), setting $x = 1, x = \frac{m}{n}$, we get
$$g(n) = (g(1))^n, \qquad (6)$$
$$g(m) = \left(g\left(\frac{m}{n}\right)\right)^n. \qquad (7)$$
From (4) we know
$$g(x) = \left(g\left(\frac{x}{2}\right)\right)^2 > 0, \text{ for any rational number } x. \qquad (8)$$
$$g(\frac{m}{n}) = (g(1))^{\frac{m}{n}}, \text{ for any positive integers } m, n. \qquad (9)$$
Also, $g(0) = \frac{f(0)}{f(0)} = 1$, so from (4) we get
$$g(-\frac{m}{n}) = \frac{g(0)}{g(\frac{m}{n})} = (g(1))^{-\frac{m}{n}}. \qquad (10)$$
From $g(0) = 1$ and equations (9), (10) we know
$$g(x) = (g(1))^x, \text{ for any rational number } x. \qquad (11)$$
From equation (11) and $g(x) = \frac{f(x)}{f(0)}$, we thus know
$$f(x) = f(0)\left(\frac{f(1)}{f(0)}\right)^x, \text{ for any rational number } x. \qquad (12)$$
Let $b = f(0) \neq 0$, $c = \frac{f(1)}{f(0)}$; equation (12) becomes $f(x) = bc^x$, for any rational number $x$. Verification shows that $f(x) = bc^x$ ($b \neq 0, c > 0$) satisfies the given condition.
Hence the desired $f(x) = bc^x$ ($b \neq 0, c > 0, b, c$ are constants).