Let $y=0$, we get
$f^{2}(x) f(0)+f^{2}(0) f(2 x)=2 f^{2}(x) f(0)$.
Thus, $f(2 x)=\frac{f^{2}(x)}{f(0)}, f(2 y)=\frac{f^{2}(y)}{f(0)}$.
Substituting equation (2) into equation (1) yields
$$f(x) f(y)(f(x) f(y)-f(0) f(x+y))=0.$$

From equation (3) and $f(x) \neq 0, f(y) \neq 0$, we have
$$f(x) f(y)-f(0) f(x+y)=0.$$

Let $g(x)=\frac{f(x)}{f(0)}$, then from equation (4) we know
$$g(x+y)=g(x) g(y),$$

and $g(x)$ is a continuous function.
From equation (5) and mathematical induction, we know
$g(n x)=(g(x))^{n}$, for any $n \in \mathbf{N}_{+}$.
Substituting $x=1, x=\frac{m}{n}$ into equation (6) gives
$g(n)=(g(1))^{n}$,
$g(m)=\left(g\left(\frac{m}{n}\right)\right)^{n}$.
From equation (5), we know
$g(x)=\left(g\left(\frac{x}{2}\right)\right)^{2}>0$, for any $x \in \mathbf{R}$.
From equations (7), (8), and (9), we know
$g\left(\frac{m}{n}\right)=(g(1))^{\frac{m}{n}}, m, n \in \mathbf{N}_{+}$.
Since $g(0)=\frac{f(0)}{f(0)}=1$, from equation (5) we get
$g\left(-\frac{m}{n}\right)=\frac{g(0)}{g\left(\frac{m}{n}\right)}=(g(1))^{-\frac{m}{n}}$.
From $g(0)=1$ and equations (10) and (11), we know
$g(x)=(g(1))^{x}$, for any $x \in \mathbf{Q}$.
From equation (2) and the continuity of $g(x)$, we know
$g(x)=(g(1))^{x}$, for any $x \in \mathbf{R}$.
(13)
From equation (13) and $g(x)=\frac{f(x)}{f(0)}$, we know
$f(x)=f(0)\left(\frac{f(1)}{f(0)}\right)^{x}$, for any $x \in \mathbf{R}$.
(14)
Let $b=f(0) \neq 0, c=\frac{f(1)}{f(0)}>0$, then equation (4) is $f(x)=b c^{x}$, for any $x \in \mathbf{R}$.
Upon verification, $f(x)=b c^{x}(b \neq 0, c>0)$ satisfies equation (1).
Therefore, the solution is $f(x)=b c^{x}(b \neq 0, c>0, b, c$ are constants $)$.