1. Given Condition and Initial Setup:
We are given a triangle $ABC$ with centroid $G$. Points $D$ and $E$ lie on segments $AB$ and $AC$ respectively, such that:
$$\frac{AB}{AD} + \frac{AC}{AE} = 3$$
We need to prove that points $D$, $G$, and $E$ are collinear.

2. Rewriting the Given Condition:
The given condition can be rewritten as:
$$\frac{AB}{AD} + \frac{AC}{AE} = 3$$
This implies:
$$AB \cdot AE + AC \cdot AD = 3 \cdot AD \cdot AE$$

3. Area Interpretation:
Multiplying both sides by $\frac{1}{2} \sin \angle CAB$, we get:
$$\frac{1}{2} AB \cdot AE \sin \angle CAB + \frac{1}{2} AC \cdot AD \sin \angle CAB = \frac{3}{2} AD \cdot AE \sin \angle CAB$$
This can be interpreted as:
$$[\triangle ADC] + [\triangle AEB] = 3 [\triangle ADE]$$
where $[ \triangle XYZ ]$ denotes the area of triangle $XYZ$.

4. Area Relationship:
Since $[\triangle ADC] + [\triangle AEB] = 3 [\triangle ADE]$, we can further deduce:
$$[\triangle EDC] + [\triangle EDB] = [\triangle ADE]$$
This implies that the sum of the areas of $\triangle EDC$ and $\triangle EDB$ equals the area of $\triangle ADE$.

5. Base and Height Consideration:
Since these triangles share the same base $DE$, we can consider the heights from points $B$, $C$, and $A$ to line $DE$. Let $R$ be the foot of the perpendicular from $C$ to $DE$, $S$ be the foot of the perpendicular from $B$ to $DE$, and $T$ be the foot of the perpendicular from $A$ to $DE$.

6. Height Relationship:
Given the area relationship, we have:
$$CR + BS = TA$$
Since $BCRS$ forms a right trapezoid and $M$ is the midpoint of $BC$, we have:
$$UM = \frac{CR + BS}{2}$$
and thus:
$$UM = 2 \cdot TA$$

7. Triangles Similarity:
Consider triangles $\triangle UG'M$ and $\triangle TG'A$, where $G' = UT \cap AM$. Since $\angle MUG' = \angle ATG' = 90^\circ$ and $\angle UG'M = \angle TG'A$, we have:
$$\triangle UG'M \sim \triangle TG'A$$
Therefore:
$$\frac{UM}{TA} = \frac{MG'}{AG'} = \frac{G'U}{G'T} = \frac{1}{2}$$
Hence, $G' = G$ because the centroid $G$ bisects the median into two segments with a ratio of $2:1$.

8. Conclusion:
Since $G'$ lies on $UT$ and $U$ and $T$ are on line $DE$, we conclude that $D$, $G$, and $E$ are collinear.

$\blacksquare$