3. As shown in Figure 1, in equilateral

triangle $\triangle A B C$, $M$ and $N$ are the midpoints of sides $A B$ and $A C$, respectively. $D$ is any point on $M N$, and the extensions of $B D$ and $C D$ intersect $A C$ and $A B$ at points $E$ and $F$, respectively. If $\frac{1}{C E}+\frac{1}{B F}=3$, then $S_{\triangle A B C}$ is ().
(A) 1
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{3 \sqrt{3}}{4}$
(D) $\frac{\sqrt{3}}{4}$
3.D.

As shown in Figure 4, draw $DS \parallel BM$ and $DT \parallel CN$ intersecting $BC$ at $S$ and $T$. Since $M$ and $N$ are the midpoints of sides $AB$ and $AC$ respectively, and $MN = \frac{1}{2} BC$. Therefore, quadrilaterals $BSDM$ and $TCND$ are both parallelograms. Hence,
$$
MD = BS, DN = TC.
$$

Thus, $ST = MN$.
$$
\text{Also, } \angle DST = \angle ABC = 60^\circ, \angle DTS = \angle ACB = 60^\circ,
$$
so, $\triangle DST$ is also an equilateral triangle.
Since $DT \parallel AC$, we have $\frac{DT}{CE} = \frac{BT}{BC}$.
Since $DS \parallel AB$, we have $\frac{DS}{BF} = \frac{CS}{CB}$.
Therefore, $\frac{1}{CE} + \frac{1}{BF} = \frac{CS + BT}{BC \cdot ST} = \frac{\frac{3}{2} BC}{BC \cdot \frac{1}{2} BC} = 3$.
Thus, $BC = 1$.
Since $\triangle ABC$ is an equilateral triangle, we have,
$$
S_{\triangle ABC} = \frac{1}{2} \times 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}.
$$