Based on the following reaction,

${\mathrm{XeO}}_6^{4-} \left(\mathrm{aq}\right) + 2{\mathrm{F}}^{-} \left(\mathrm{aq}\right) + 6{\mathrm{H}}^{+}\left(\mathrm{aq}\right) \stackrel{ }{\to } {\mathrm{XeO}}_3\left(\mathrm{g}\right) + {\mathrm{F}}_2\left(\mathrm{g}\right) + 3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \left(∆{\mathrm{G}}^0<0\right)$

It can be concluded that 

$\stackrel{+8}{\mathrm{Xe}}{\mathrm{O}}_6^{-4} + {\mathrm{F}}^{-} \to \stackrel{+6}{\mathrm{Xe}}{\mathrm{O}}_3 +{\stackrel{0}{\mathrm{F}}}_2$

Since, $∆G^0 < 0,$ hence it is spontaneous in forward direction. Oxidation number of Xe decreases, hence, it is an oxidizing agent; and oxidation number of F increases, hence it is a reducing agent.