I found a pretty "forceful" method using modular artihmetics. First consider equation $ f^2 \equiv f $. Now we can try for numbers trought 0-9 and see that only 0,1,5,6 meet the requirements. Now nosider $(ef)$, where $f$ can only be 0,1,5,6 and $e$ can be 0-9. After that we get that the number $ (ef)=25$ is good. Now we do the same thing for $(def)$ and get $(def) =625$, which squared gives: 390625. I know that it ain't pretty but took me like 15mins of squaring stuff by hand.

This looks nice. Let $x:=\overline{abc},y:=\overline{def}$ then $1000x+y=y^2 \iff y(y-1)=1000x$. Then $y\equiv 0,1 (8)$ since exactly one of the numbers $y,y-1$ is even. In the same way $y\equiv 0,1(125)$. Combining them with CRT gives $y\equiv 0,1,625,376$. I guess we'll have to discard the first two and $625^2=390625,376^2=141376$. Basically we determined the idempotent elements in the ring $Z/1000$.