Example 4 As shown in Figure 3, in the acute $\triangle A B C$, $D 、 E 、 F$ are the trisection points on sides $A B 、 B C 、 C A$, respectively, and $P 、 Q 、 R$ are the intersections of the three medians of $\triangle A D F 、 \triangle B D E 、 \triangle C E F$, respectively. Find:
(1) $S_{\triangle D E F}: S_{\triangle A B C}$;
(2) $S_{\triangle P D F}: S_{\triangle A D F}$;
(3) $S_{\text {hexagon PDQERF }}: S_{\triangle A B C}$.
(1) As shown in Figure 3, draw $D G \perp B C$ at point $G$, and draw $A H \perp B C$ at point $H$.
Then $D G / / A H$.
Therefore, $\triangle B D G \backsim \triangle B A H$.
Since $\frac{B D}{B A}=\frac{1}{3}, B E=\frac{2}{3} B C$, we have
$D G=\frac{1}{3} A H, S_{\triangle B D E}=\frac{2}{9} S_{\triangle A B C}$.
Similarly, $S_{\triangle A D F}=S_{\triangle C E F}=\frac{2}{9} S_{\triangle A B C}$.
Thus, $S_{\triangle D E F}=S_{\triangle I B C}-S_{\triangle A D F}-S_{\triangle C E F}-S_{\triangle A D E}$ $=\frac{1}{3} S_{\triangle A B C}$.
(2) As shown in Figure 3, extend $D P$ to intersect $A F$ at point $M$.
Since $P$ is the intersection of the three medians of $\triangle A D F$, $M$ is the midpoint of $A F$, and $D P=\frac{2}{3} D M$.

Draw perpendiculars from points $P$ and $M$ to $D F$, with feet at $K$ and $S$ respectively.
Then $\triangle D K P \backsim \triangle D S M$, with a similarity ratio of $2: 3$.
Therefore, $K P=\frac{2}{3} S M, S_{\triangle P D F}=\frac{2}{3} S_{\triangle M D F}$.
Since $S_{\triangle M D F}=\frac{1}{2} S_{\triangle I D F}$, we get
$S_{\triangle P I F}=\frac{1}{3} S_{\triangle A D F}$.
(3) From (2), we know
$S_{\triangle Q D E}=\frac{1}{3} S_{\triangle B D E}, S_{\triangle R E F}=\frac{1}{3} S_{\triangle C E F}$.
Therefore, $S_{\triangle P D F}=S_{\triangle Q D E}=S_{\triangle R E F}=\frac{1}{27} S_{\triangle A B C}$.
Thus, $S_{\text {mice } \ln P Q E R F}$
$$
\begin{array}{l}
=S_{\triangle D E F}+S_{\triangle P D F}+S_{\triangle O D E}+S_{\triangle R E F} \\
=\frac{5}{9} S_{\triangle A H C} .
\end{array}
$$