$(1)$证明：在$\triangle ABC$中，点$D$在边$AB$上，点$E$、点$F$在边$AC$上，且$DE$∥$BC$，$\frac{AF}{FE}=\frac{AE}{EC}$，
$\therefore \frac{AD}{DB}=\frac{AE}{EC}$，
$\therefore \frac{AD}{DB}=\frac{AF}{FE}$，
$\therefore \frac{AD}{AB}=\frac{AF}{AE}$，
$\because \angle A=\angle A$，
$\therefore \triangle ADF$∽$\triangle ABE$，
$\therefore \angle ADF=\angle ABE$，
$\therefore DF$∥$BE$；
$(2)$$\because \frac{AF}{FE}=\frac{AE}{EC}$，$EF=2AF$，
$\therefore \frac{AF}{2AF}=\frac{AE}{EC}=\frac{1}{2}$，
$\therefore EC=2AE=2\left(AF+EF\right)=6AF$，
$\because DF$∥$BE$，
$\therefore \angle FDE=\angle BED$，
$\because DE$∥$BC$，
$\therefore \angle CBE=\angle BED$，$\angle DEF=\angle C$，
$\therefore \angle CBE=\angle FDE$，
$\therefore \triangle DEF$∽$\triangle BCE$，
$\therefore \frac{{C}_{△DFE}}{{C}_{△BEC}}=\frac{EF}{CE}=\frac{2AF}{6AF}=\frac{1}{3}$.

$(1)$证明：在$\triangle ABC$中，点$D$在边$AB$上，点$E$、点$F$在边$AC$上，且$DE$∥$BC$，$\frac{AF}{FE}=\frac{AE}{EC}$，
$\therefore \frac{AD}{DB}=\frac{AE}{EC}$，
$\therefore \frac{AD}{DB}=\frac{AF}{FE}$，
$\therefore \frac{AD}{AB}=\frac{AF}{AE}$，
$\because \angle A=\angle A$，
$\therefore \triangle ADF$∽$\triangle ABE$，
$\therefore \angle ADF=\angle ABE$，
$\therefore DF$∥$BE$；
$(2)$$\because \frac{AF}{FE}=\frac{AE}{EC}$，$EF=2AF$，
$\therefore \frac{AF}{2AF}=\frac{AE}{EC}=\frac{1}{2}$，
$\therefore EC=2AE=2\left(AF+EF\right)=6AF$，
$\because DF$∥$BE$，
$\therefore \angle FDE=\angle BED$，
$\because DE$∥$BC$，
$\therefore \angle CBE=\angle BED$，$\angle DEF=\angle C$，
$\therefore \angle CBE=\angle FDE$，
$\therefore \triangle DEF$∽$\triangle BCE$，
$\therefore \frac{{C}_{△DFE}}{{C}_{△BEC}}=\frac{EF}{CE}=\frac{2AF}{6AF}=\frac{1}{3}$.