Put $x = 0$, we get $2f(0)f(y) - f(0) = 0$, if $f(0) \neq 0$ then $f(y) = 1/2$ for all $y$, which does not satisfy. Thus $f(0) = 0$.

Put $y = 0$, we get $2f(x)^2 - f(x^2) = x/2 \cdot f(2x)$ then plugging back $2f(x) \cdot f(x+y) = 2f(x)^2 + 2x \cdot f(f(y))$ so
$$f(x) \cdot f(x+y) = f(x)^2 + x \cdot f(f(y)). \qquad (\diamond)$$
Note that $f(-x)^2 = x \cdot f(f(x))$. In $(\diamond)$, take $y = -2x$ then
$$f(x) \cdot f(-x) = f(x)^2 + x \cdot f(f(-2x)).$$
Note that $-2x \cdot f(f(-2x)) = f(2x)^2$, so $x \cdot f(f(-2x)) = -f(2x)^2/2$. Thus
$$f(x) \cdot f(-x) = f(x)^2 - f(2x)^2/2$$
or
$$-2f(x)f(-x) + 2f(x)^2 = f(2x)^2.$$
In $(\diamond)$, take $x = y \neq 0$ then
$$f(x) \cdot f(2x) = f(x)^2 + x \cdot f(f(x))$$
or
$$f(2x) = f(x) + x \cdot f(f(x))/f(x) = f(x) + f(-x)^2/f(x).$$
So $(f(x) + f(-x)^2/f(x))^2 = 2f(x)^2 - 2f(x) \cdot f(-x)$. Dividing both sides by $f(-x)^2$ and put $k = f(x)/f(-x)$ then
$$(k + 1/k)^2 = 2k^2 - 2k \longleftrightarrow k^2 - 2k = 2 + 1/k^2.$$
Changing the sign $x \to -x$ then we get $(k + 1/k)^2 = 2/k^2 - 2/k$ combining with above, we get $k = -1$. Then $f(x)$ is odd. Now we get $f(x)^2 = x \cdot f(f(x))$. Put $y \to -y$, we get
$$f(x) \cdot f(x - y) = f(x)^2 + x \cdot f(f(-y)),$$
$$f(x)[f(x + y) + f(x - y)] = 2f(x)^2.$$
Consider $x \neq 0$ then $f(x + y) + f(x - y) = 2f(x)$ using $f(0) = 0$, this implies that $f(x)$ is additive. Note that
$$2f(x)^2 - f(x^2) = x/2 \cdot f(2x) = x \cdot f(x) \iff 2f(x)^2 = f(x^2) + xf(x).$$
Plugging $x \to x + 1$, it is easily to get $f(x) = ax$ and putting back to the original, $a = 1$. There are two solutions: $f(x) = 0, f(x) = x$. $\square$