Solution:
The only solutions are $f(x)=x$ for all rational $x$ and $f(x)=-x$ for all rational $x$. Both of these readily check out.

Setting $y=x$ yields $f(3 f(x))=3 x$ for all rational $x$. Now replacing $x$ by $3 f(x)$, we find that
$$f(9 x)=f(3 f(3 f(x)))=3[3 f(x)]=9 f(x),$$
for all rational $x$. Setting $x=0$ yields $f(0)=9 f(0)$, whence $f(0)=0$.

Setting $x=0$ in the given functional equation yields $f(f(y))=y$ for all rational $y$. Thus $f$ is one-one onto. Applying $f$ to the functional equation yields that
$$2 f(x)+f(y)=f(2 x+y)$$
for every rational pair $(x, y)$.

Setting $y=0$ in the functional equation yields $f(2 f(x))=2 x$, whence $2 f(x)=f(2 x)$. Therefore $f(2 x)+f(y)=f(2 x+y)$ for each rational pair $(x, y)$, so that
$$f(u+v)=f(u)+f(v)$$
for each rational pair $(u, v)$.

Since $0=f(0)=f(-1)+f(1)$, $f(-1)=-f(1)$. By induction, it can be established that for each integer $n$ and rational $x$, $f(n x)=n f(x)$. If $k=f(1)$, we can establish from this that $f(n)=n k$, $f(1 / n)=k / n$ and $f(m / n)=m k / n$ for each integer pair $(m, n)$. Thus $f(x)=k x$ for all rational $x$. Since $f(f(x))=x$, we must have $k^2=1$. Hence $f(x)=x$ or $f(x)=-x$. These check out.

Solution:
In the functional equation, let
$$x=y=2 f(z)+f(w)$$
to obtain $f(x)=f(y)=2 z+w$ and
$$f(6 z+3 w)=6 f(z)+3 f(w)$$
for all rational pairs $(z, w)$. Set $(z, w)=(0,0)$ to obtain $f(0)=0$, $w=0$ to obtain $f(6 z)=6 f(z)$ and $z=0$ to obtain $f(3 w)=3 f(w)$ for all rationals $z$ and $w$. Hence $f(6 z+3 w)=f(6 z)+f(3 w)$. Replacing $(6 z, 3 w)$ by $(u, v)$ yields
$$f(u+v)=f(u)+f(v)$$
for all rational pairs $(u, v)$. Hence $f(x)=k x$ where $k=f(1)$ for all rational $x$. Substitution of this into the functional equation with $(x, y)=(1,1)$ leads to $3=f(3 f(1))=f(3 k)=3 k^2$, so that $k= \pm 1$. It can be checked that both $f(x) \equiv x$ and $f(x) \equiv -x$ satisfy the equation.