Rather than show that AF and BD intersect at G, it's easier to show that AF and BD (separately), pass through the midpoint of CE.
Dealing with AF first, suppose that AF intercects CE at P. We then have to show that P is the midpoint of CE.
Let $\displaystyle \underline{AP}=\lambda\underline{AF},$ and let $\displaystyle \underline{PE}=\mu\underline{CE}$ where $\displaystyle \lambda \text { and } \mu$ are constants, each less than 1.
From the triangle ABF we have
$\displaystyle \underline{AF} = \underline{AB}+ \underline{BF} = \underline{AB}+(3/5)\underline{BC},$
and from BEC,
$\displaystyle \underline{CE} = \underline{CB}+ \underline{BE} = \underline{CB}+(1/3)\underline{BA}.$
From the triangle APE,
$\displaystyle \underline{AE} = \underline{AP}+ \underline{PE} = \lambda\underline{AF}+\mu\underline{CE},$
from which, substituting from the earlier equations,
$\displaystyle (2/3)\underline{AB}=\lambda\{ \underline{AB}+(3/5)\underline{BC} \}+\mu \{\underline{CB}+(1/3)\underline{BA} \}.$
Collecting terms,
$\displaystyle \left( \frac{2}{3}-\lambda+\frac{\mu}{3} \right) \underline{AB}= \left( \frac{3\lambda}{5}-\mu \right) \underline{BC}.$
The vectors $\displaystyle \underline{AB} \text{ and } \underline{BC}$ are not parallel, so for equality it's necessary that the two expressions in brackets are each zero.
Solving simultaneously produces $\displaystyle \lambda = 5/6 \text{ and } \mu = 1/2, $ the latter indicating that P is the midpoint of CE and hence coincident with G.
Geometrically, the result is obvious. Joining the two points between B and F to the two points on AB gets you two lines parallel with AF, and joining the point between F and C to the midpoint of AC gets you a fourth line also parallel to AF.
The lines are equally spaced in which case CE will be divided into four parts all of equal length, two to the left of P and two to the right. Therefore, P will be the midpoint of CE.
The vector routine can be used to show that BD also passes through the point G.