First Solution. Plugging $(x, y) = (1, yf(z) + f(z)f(y))$ yields
$$\begin{align*} & f(C(yf(z) + f(z)f(y) + f(yf(z) + f(z)f(y)))) \\ &= f(C(yf(z) + zf(y) + f(y)f(z) + f(zy)) \\ &= 2f(yf(z) + f(z)f(y)) = 2zf(y) + 2f(zy) \end{align*}$$

$$zf(y) + f(zy) = yf(z) + f(yz).$$
Hence, $f(y) = Cy$ for some constant $C$. Putting $f(x) = Cx$ in the original equation yields $f(x) = 0, f(x) = x, f(x) = -2x$.

Second Solution. As in the first solution, $f(1) = a$. Assume that $f$ is not constant, then, $f((1+a)f(x)) = ax + f(x)$ yielding to the fact that $f$ is injective. Further, $f(a(y+f(y))) = 2f(y)$. It follows that if $r+f(r)=s+f(s)$ then $r=s$. Hence, $f(a^2+a) = 2a$ and $f((1+a)f(a^2+a)) = f(2a^2+2a) = a(a^2+a) + f(a^2+a) = a^3 + a^2 + 2a$, finally, $f(a(a^2+a+f(a^2+a))) = f(a^3+3a^2) = 2f(a^2+a) = 4a$. Notice that $a^3+3a^2+4a = a^3+a^2+2a+2a^2+2a$. It follows that $f(2a^2+2a) + 2a^2 + 2a = f(a^3+3a^2) + a^3 + 3a^2$. Hence, $2a^2 + 2a = a^3 + 3a^2$. Thus, $a \in \{-2, 1\}$.
If $a=1$ then $f(2f(x)) = x+f(x)$ letting $x=2f(z)$ to obtain $f(y(z+f(z)) + (z+f(z))f(y)) = f(yz+zf(z)+zf(y)+f(y)f(z)) = 2f(z)f(y)+f(2yf(z))$, interchanging $y,z$ to obtain $f(2zf(y)) = f(2yf(z))$ since $f$ is injective, it follows that $zf(z) = zf(y)$ and hence $f(x) = x$.
If $a = -2$ then $f(-f(x)) = f(x) - 2x$ and $f(-2f(x)-2x) = 2f(x)$ putting $-f(x)$ instead of $x$ in the second equation to obtain $f(4x) = 2f(x) - 4x$. Plugging $-2(x+f(x))$ instead of $x$ in the second equation to obtain $f(4x) = 4f(x)$. Hence, $f(x) = -2x$. ■