You first need to consider if $[x]_3\mapsto [2x]_6$ is well-defined. Say that $[x]_3 = [y]_3$ then $x\equiv y(\bmod 3)$, so $2x\equiv 2y(\bmod 6)$ and hence $[2x]_6 = [2y]_6$ which means the mapping is well-defined. Now what about $[x]_3\mapsto [4x]_3$? Try to argue that this is also well-defined. To show it is a ring homorphism (depending on how you define "ring homomorphism") you just need to show $f(x+y) = f(x)f(y)$ and $f(xy) = f(x)f(y)$. Notice $f( [x]_3) + f([y]_3)= [2x]_6 + [2y]_6 = [2(x+y)]_6 = f([x+y]_3)$. Also, $f([x]_3)f([y]_3) = [2x]_6\cdot [2y]_6 = [4xy]_6 \not = [2xy]_6 = f([xy]_3)$. Thus, it is not a homorphism, you try doing the second case.