jmc

Triangles $AXC$ and $ABC$ share an altitude from $C$, so $$\frac{[AXC]}{[ABC]}=\frac{AX}{AB}=\frac{4}{7}.$$ Therefore, $[AXC]=\frac{4}{7}[ABC]$.

Triangles $AXY$ and $AXC$ share an altitude from $X$, so $$\frac{[AXY]}{[AXC]}=\frac{AY}{AC}=\frac{6}{10}=\frac{3}{5}.$$ Therefore, $[AXY]=\frac{3}{5}[AXC]$, so $$[AXY]=\frac{3}{5}[AXC]=\frac{3}{5}\cdot \frac{4}{7}[ABC]=\frac{12}{35}[ABC],$$ which means that the desired ratio is $\boxed{\frac{12}{35}}$.