In the following balanced reaction, 

$\mathrm{X}{\mathrm{MnO}}_4^{-}+{\mathrm{YC}}_2{\mathrm{O}}_4^{2-}+{\mathrm{ZH}}^{+} \rightleftharpoons \mathrm{X}{\mathrm{Mn}}^{2+}+2{\mathrm{YCO}}_2+\frac{\mathrm{Z}}{2}{\mathrm{H}}_2\mathrm{O}$
values of X, Y and Z respectively are 

$\mathrm{X}{\mathrm{MnO}}_4^{-}+{\mathrm{YC}}_2{\mathrm{O}}_4^{--}+{\mathrm{ZH}}^{+}\leftrightharpoons \mathrm{X}{\mathrm{Mn}}^{++}+2\mathrm{Y}{\mathrm{CO}}_2+\frac{\mathrm{Z}}{2}{\mathrm{H}}_2\mathrm{O}$
First half reaction 
${\mathrm{MnO}}_4^{-}⟶{\mathrm{Mn}}^{++} ...\left(\mathrm{i}\right)$

On balancing 
${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}⟶{\mathrm{Mn}}^{++}+4{\mathrm{H}}_2\mathrm{O} ...\left(\mathrm{ii}\right)$
Second half reaction 
${\mathrm{C}}_2{\mathrm{O}}_4^{--}⟶2{\mathrm{CO}}_2 .....\left(\mathrm{iii}\right)$
On balancing 
${\mathrm{C}}_2{\mathrm{O}}_4^{--}⟶2{\mathrm{CO}}_2+2{\mathrm{e}}^{-} ....\left(\mathrm{iv}\right)$
On balancing 
${\mathrm{C}}_2{\mathrm{O}}_4^{--}⟶2{\mathrm{CO}}_2+2{\mathrm{e}}^{-} ...\left(\mathrm{iv}\right)$
On multiplying eqn. (ii) by 5 and (iv) by 2 and then adding we get 
$2{\mathrm{MnO}}_4^{-}+5{\mathrm{C}}_2{\mathrm{O}}_4^{--}+16{\mathrm{H}}^{+}⟶ 2{\mathrm{Mn}}^{++}+10{\mathrm{CO}}_2+8{\mathrm{H}}_2\mathrm{O}$