3. As shown in Figure 4, AM is the angle bisector of $\triangle A B C$, and $D, E$ are points on sides $A B, A C$ respectively. $D E$ intersects $A M$ at point $F$. If $A D=1, A E=2, B D$ $=3, E C=4$, then $\frac{A F}{A M}=$ $\qquad$
3. $\frac{5}{18}$.

Given that $A M$ bisects $\angle B A C$, we have
$$
\frac{D F}{F E}=\frac{A D}{A E}=\frac{1}{2}, \quad \frac{B M}{M C}=\frac{A B}{A C}=\frac{2}{3} \text {. }
$$

Therefore, $S_{\triangle N F F}=\frac{1}{3} S_{\triangle A D E}, S_{\triangle B B Y}=\frac{2}{5} S_{\triangle A B C}$, which means
$$
\begin{array}{l}
\frac{S_{\triangle W F F}}{S_{\triangle A B M}}=\frac{5}{6} \cdot \frac{S_{\triangle A B E}}{S_{\triangle A B C}} . \\
\text { Also, } \frac{S_{\triangle \triangle D F}}{S_{\triangle A B E}}=\frac{\frac{1}{2} A D \cdot A F \sin \angle D A F}{\frac{1}{2} A B \cdot A M \sin \angle B A M}=\frac{1}{4} \cdot \frac{A F}{A M},
\end{array}
$$
$\frac{S_{\triangle A Q E}}{S_{\triangle A B C}}=\frac{1}{12}$, hence
$$
\frac{1}{4} \cdot \frac{A F}{A M}=\frac{5}{6} \times \frac{1}{12}, \quad \frac{A F}{A M}=\frac{5}{18} .
$$