Proof that each of the three cevians is divided in the ratio $1:3:3$ Points $D,E,F$ are the first trisection points of $BC,CA,AB$ respectively. Let $[ABC]$ denotes the area of triangle $ABC$. If $[ABC]=1$, find $[GHI]$, the area of the shaded triangle. (the two images are from "The Art and Craft of Problem Solving" -Paul Zeitz)


attempt:
Let us look at the image below first

We know $AF:FB = 1:2$ ,this implies $[FIA]=x \implies [FBI]=2x$. Similarly $[DIB]=2y \implies [CID] = 4y$
Because $[CAF]:[CFB]=1:2$ and $[CFB]=6y+2x$, then $[CAF]=3y+x$ and $[CIA]=3y$.
We know the ratio $AF:AB = 1:3$, so ratio $[CAF]:[CFB]=1:2$. 
Why at some point we can conclude that each of the three cevians $AD, EB, CF$, are divided by the ratio $1:3:3$?

We may and do assume that the area $[ABC]$ is one.
Following the arguments of the OP, we already have the following information regarding the areas cut / separated from $[ABC]$ by two cevians, $AF$ and $BD$:

In particular we have $2:1=CD:DB=\color{red}{7y}:\color{blue}{3x+2y}$,  which implies $7y=2(3x+2y)$, i.e. $y=2x$. Then we can immediately express the piece $x$ in terms of the total area 
$$
1=[ABC]=\color{red}{7y}+\color{blue}{3x+2y}=14x+3x+4x=21x
\ ,
$$
so $x=[AIF]=1/21$, and $y=2/21$.
By the same argument, and/or by cylcic shift in the notation, we have 
$$
[AIF]=[BGD]=[CHE]=\frac 1{21}\ .
$$
We also have so far $3y=[AIC]=6/21$. We remove the area of the triangle $\Delta CHE$ from this quantity, thus getting by symmetry the following quadrilateral areas:
$$
[AIHE]=[BGIF]=[CHGD]=\frac6{21}-\frac 1{21}=\frac 5{21}\ .
$$
For the triangle in the middle, the remained area is
$$
[GHI]=\frac{21-1-1-1-5-5-5}{21}=\frac 3{21}=\frac 17\ .
$$
In a picture:


Note: The above computed the area of the "inner" triangle $\Delta GHI$ without knowing the proportion $AI:IG:GD$. A posteriori, if we really need it, we can for instance compute $AI:ID=[CAI]:[CID]=3y:4y=3:4$, $AG:GD=[AGC]:[GDC]=6:1$.
Note: In such cases much of the effort is reduced by using directly the Theorem of Menelaus. For instance, applying it for $\Delta ABD$ with the line intersecting the sides in $F,C,I$, we can write immediately:
$$
\frac{FA}{FB}\cdot\frac{CB}{CD}\cdot\frac{ID}{IA}=1\ .
$$
So $AI:ID=3:4$.
