3. As shown in the figure, it is known that the area of $\triangle A B C$ is 1, and $\frac{B D}{D C}=\frac{C E}{E F}=\frac{A F}{F D}=\frac{1}{2}$. Therefore, the area of $\triangle D E F$ is
3. $\frac{8}{27}$.

Since the heights of $\triangle D E F$ and $\triangle D C F$ are equal, and $\frac{F E}{F C}=\frac{2}{3}$, then
$$
\frac{S_{\triangle D E F}}{S_{\triangle D C F}}=\frac{2}{3} \text {. }
$$

Similarly, $\frac{S_{\triangle D C F}}{S_{\triangle D C A}}=\frac{2}{3}, \frac{S_{\triangle D C A}}{S_{\triangle A B C}}=\frac{2}{3}$.
Multiplying the three equations yields $S_{\triangle D E F}=\frac{8}{27}$.