$CE \cdot CA=\tfrac{1}{3}CA \cdot CA=CD \cdot CB=\tfrac{2}{3}CB \cdot CB$ $\Longrightarrow$ $CA^2=2 \cdot CB^2$ $\Longrightarrow$ $\tfrac{CA}{CB}=\sqrt{2}$ $\Longrightarrow$ ratio of distances from $C$ to $A,B$ is constant $\Longrightarrow$ $C$ moves on the Apollonius circle $\mathcal{O}$ of $\overline{AB}$ referred to that ratio. Let $P \equiv AD \cap BE$ and $F \equiv CP \cap AB.$ Using Ceva's theorem for the concurrent cevians $AD,BE,CF$ and Menelaus' theorem for $\triangle CBF$ cut by $APD,$ we get $\frac{AF}{FB}=\frac{DC}{BD} \cdot \frac{EA}{CE}=2 \cdot 2=4 \ , \ \frac{PF}{CP}=\frac{AF}{AB} \cdot \frac{BD}{DC}=\frac{4}{5} \cdot \frac{1}{2}=\frac{2}{5} \Longrightarrow \frac{FP}{FC}=\frac{2}{7}.$ Hence, $F$ is fixed and the ratio $\tfrac{FP}{FC}$ is constant $\Longrightarrow$ locus of $P$ is the homothetic circle of $\mathcal{O}$ under homothety with center $F$ and ratio $\tfrac{2}{7}.$

it's easy by Appalonious circle. Locus is circle.