Put $x=y$ get $f(2x)=4f(x)$ Now by induction $f(2^nx)=2^{2n}f(x)$ Let $g(x)=\frac {f(x)}{x}$, we'll prove $g(nx)=ng(x)$ for all natural number $n$. It's true for $n=2^m$ We've $g(x+y)\leq g(x)+g(y)$ so $g(nx)\leq ng(x)$ Choosing $2^{m-1}\leq n<2^m$ Now $2^mg(x)=g(2^mx)\leq g(nx)+g(2^mx-nx)\leq ng(x)+(2^m-n)g(x)=2^mg(x)$ So now we've $g(nx)=ng(x)$ It's easy to see $g(x)$ is a non increasing function. $2(f(x)+f(2x))\leq f(3x)$ from here easily we'll get $g(x)\leq 0$ for all $x\geq 0$. From condition $g(nx)=ng(x)$ we have $g(x)=xg(1)$ for all positive rational $x$ Now as it non increasing so using a well known theorem we've $g(x)=xg(1)$ for all $x\geq 0$ So $f(x)=f(1)x^2$. ($f(1)\leq 0$)

Beautiful solution. May be to show that the function g is decreasing