jmc

$7@4=7\cdot 4-2\cdot 7=14$ and $4@7=4\cdot 7-2\cdot 4=20$, so $(7@4)-(4@7)=14-20=\boxed{-6}$. Another way to solve this problem is to realize that the expression $(7@4)-(4@7)$ is of the form $(x@y)-(y@x)=xy-2x-yx+2y=-2x+2y$, so the expression is just equal to $-2\cdot 7+2\cdot 4=\boxed{-6}$.