This is a classical one. The answer is $f(x)\equiv 3x$. It is easy to see that it satisfies the equation. Now, define $a_n=f^n(x)$ for $n\geq 0$ where we take $f^0(x)=x$, so we have \[x\rightarrow f^n(x) : a_{n+2}+a_{n+1}=12a_n\]and this recurrence relation has the solution $a_n=c_1\cdot 3^n+c_2 \cdot (-4)^n$ since we know that $f(x)\geq 0$ for all positive reals, the term $c_2 \cdot (-4)^n$ must be $0$ as otherwise, for some very large $n$, $f$ will be negative. Thus, $a_n=c\cdot 3^n$ and $f(c\cdot 3^n)=f(a_n)=a_{n+1}=c\cdot 3^{n+1}$. Since $c\cdot 3^{n+1}$ is surjective ($c$ is an arbitrary positive real number). We have $f(x)=3x$ as desired. $\blacksquare$