6. As shown in the figure, in $\triangle A B C$, points $D, E, F$ are on $B C, C A, A B$ respectively, and $C D=\frac{3}{5} B C, E C=\frac{1}{2} A C$, $A F=\frac{1}{3} A B$. Let $P$ be a point inside quadrilateral $A E D F$ (point $P$ is not on the boundary). If $\overrightarrow{D P}=-\frac{1}{3} \overrightarrow{D C}+x \overrightarrow{D E}$, then the range of the real number $x$ is . $\qquad$
Given that the midpoint of $BD$ is $G$, then $\overrightarrow{D G}=-\frac{1}{3} \overrightarrow{D C}$. Draw a line through $G$ parallel to $DE$ intersecting $DF$ and $AE$ at $M$ and $N$, respectively. Thus, point $P$ lies on segment $MN$. Since $\overrightarrow{D M}=\overrightarrow{D G}+\overrightarrow{G M}$ and $\overrightarrow{D N}=\overrightarrow{D G}+\overrightarrow{G N}$, it follows that $x \overrightarrow{D E}$ lies between $\overrightarrow{G M}$ and $\overrightarrow{G N}$. Noting that $\frac{C E}{E N}=3 \Rightarrow \frac{A N}{N C}=\frac{2}{4}=\frac{1}{2}$ $\Rightarrow \frac{A N}{A C}=\frac{1}{3} \Rightarrow F N \parallel B C$. Therefore, $\frac{F N}{B C}=\frac{1}{3} \Rightarrow F N=\frac{1}{3} B C=\frac{5}{3} G D \Rightarrow \frac{G M}{M N}=\frac{G D}{F N}=\frac{3}{5}$ $\Rightarrow G M=\frac{3}{8} G N=\frac{3}{8} \cdot \frac{4}{3} D E=\frac{1}{2} D E \Rightarrow \frac{1}{2}<x<\frac{4}{3}$. Thus, the range of the real number $x$ is $\left(\frac{1}{2}, \frac{4}{3}\right)$.