jmc

Since $△CDE$ has a right angle at $D$ and $\angle C=60^{\circ },$ we can let $CD=x,$ $DE=x\sqrt{3},$ and $CE=2x$ for some positive $x.$ Note that $△AEF\cong △CDE,$ because $\angle AEF=180^{\circ }-\angle DEF-\angle CED=180^{\circ }-60^{\circ }-30^{\circ }=90^{\circ },$ $\angle EAF=60^{\circ },$ and $EF=DE.$ Then $AE=CD=x,$ so the side length of $△ABC$ is $AC=AE+EC=2x+x=3x.$

Finally, the ratio of the areas of the triangles is the square of the ratio of the side lengths: $${\left(\frac{DE}{AC}\right)}^2={\left(\frac{x\sqrt{3}}{3x}\right)}^2=\boxed{\frac{1}{3}}.$$