Sorry Melody, but that is wrong.
It's a perfectly acceptable exercise in vector addition.
a and b are vectors, they each have an magnitude, (unknown), and a direction. The direction of a is parallel to CA, in that direction and b is parallel to and in the direction of CB.
Vectors obey a triangle law of addition.
In the original diagram, put a line between C and X, then the vector CX will equal the sum of the vectors CA and AX.
\(\displaystyle \underline{CX}=\underline{CA}+\underline{AX}\)
Similarly,
\(\displaystyle \underline{CX}=\underline{CB}+\underline{BX}\) ,
so long as you start and end at the same points, you can visit any other point(s) on the way.
The law can be extended to more than one intermediate point.
For example, put a line between Y and X, and you can say that \(\displaystyle \underline{BX}=\underline{BY}+\underline{YX}\),
so \(\displaystyle \underline{CX}=\underline{CB}+\underline{BY}+\underline{YX}\).
So long as you start and end at the same points, ... .
Onto the actual question.
We are told that \(\displaystyle \underline{AX}:\underline{XB}=1:2 \text{ , so, }2\underline{AX}=\underline{XB}\).
Using the triangle law,
\(\displaystyle \underline{CX}=\underline{CA}+\underline{AX}=3\underline{a}+\underline{AX}\),
and
\(\displaystyle \underline{CX}=\underline{CB}+\underline{BX}=\underline{CB}-\underline{XB}=6\underline{b}-2\underline{AX}\).
Eliminate the \(\displaystyle \underline{AX}\) between the last two equations, (twice the first plus the second), and we have
\(\displaystyle 3\underline{CX}=6\underline{a}+6\underline{b}\), so \(\displaystyle \underline{CX}=2(\underline{a}+\underline{b})\).
Using the triangle law again,
\(\displaystyle \underline{CY}=\underline{CB}+ \underline{BY}= 6\underline{b}+(5\underline{a}-\underline{b})=5(\underline{a}+\underline{b})\).
So, \(\displaystyle \underline{CX}= 2(\underline{a}+\underline{b})=\frac{2}{5}.5(\underline{a}+\underline{b})=\frac{2}{5}\text{ .}\underline{CY}\),
as required.
Tiggsy