Since $f$ is increasing, $h(x)=f(x)+x$ is then injective. The first equation implies $h(f(x)+2g(x)+3f(y))=3f(x)+3g(x)+3f(y)+3g(y)$. If we switch $x,y$, we get $f(x)+2g(x)+3f(y)=f(y)+2g(y)+3f(x)$, or $f(x)-g(x)=c$ for some constant $c$. Let $x=y$ in the second equation, we see $g$ can take any value greater than $c$. Hence $f$ can take any value greater than $2c$ and $c\geq0$. Apply $f(x)-g(x)=c$ to the first equation, we have $f(3f(x)+3f(y)-2c)=3f(x)+3f(y)-4c$. Or $f(x)=x-2c,\forall x>10c$. In addition, $g(x)=x-3c,\forall x>10c$. Let $x,y$ be sufficiently large in the second equation, we can show $c=0$. So the solution is $f(x)=g(x)=x$.

very nice and beautiful solution!