jmc

First observe that if one vertex of a triangle is moved directly toward another vertex so as to shrink one side length of the triangle by a factor of $k$, then the area of the triangle is also shrinked by $k$. To see this, think of the side that is shrinking as the base in the equation $\text{area}=\frac{1}{2}(\text{base})(\text{height})$.

Use brackets to denote area; for example, $[ABC]$ refers to the area of triangle $ABC$. We have $$[DBE]=\frac{1}{3}[DBC]=\frac{1}{3}\left(\frac{2}{3}[ABC]\right)=\frac{2}{9}[ABC].$$ Similarly, $[ADF]=[CFE]=\frac{2}{9}[ABC]$. Therefore, $$\begin{array}{rl}[DEF]& =[ABC]-[ADF]-[CFE]-[DBE]\\ & =\left(1-\frac{2}{9}-\frac{2}{9}-\frac{2}{9}\right)[ABC]\\ & =\frac{1}{3}[ABC],\end{array}$$ so $[DEF]/[ABC]=\boxed{\frac{1}{3}}$.