jmc

We have $f(2)=2(2)+3=7$ and $g(2)=3(2)-2=4$, so $$\frac{f(g(f(2)))}{g(f(g(2)))}=\frac{f(g(7))}{g(f(4))}.$$ We then have $g(7)=3(7)-2=19$ and $f(4)=2(4)+3=11$, so we have $$\frac{f(g(7))}{g(f(4))}=\frac{f(19)}{g(11)}=\frac{2(19)+3}{3(11)-2}=\boxed{\frac{41}{31}}.$$