First off, you need to recognize this reaction as a REDOX reaction.
Then split the equation into its separate half-reactions.
MnO_(4)^(-)(aq)rarrMn^(2+)(aq)
H_2C_2O_4(aq)rarrCO_2(g)
Balance everything except the oxygen and hydrogen.
MnO_(4)^(-)(aq)rarrMn^(2+)(aq)
H_2C_2O_4(aq)rarr2CO_2(g)
Now balance the oxygen by adding H_2O(l).
MnO_(4)^(-)(aq)rarrMn^(2+)(aq)+4H_2O(l)
H_2C_2O_4(aq)rarr2CO_2(g)
Since this is in an acidic solution, the hydrogen can be balanced by adding H^(+)(aq).
MnO_(4)^(-)(aq)+8H^+(aq)rarrMn^(2+)(aq)+4H_2O(l)
H_2C_2O_4(aq)rarr2CO_2(g)+2H^+(aq)
Add electrons (e^-) in order to balance the overall charge of the two sides of either half-reaction.
MnO_(4)^(-)(aq)+8H^+(aq)+5e^(-)rarrMn^(2+)(aq)+4H_2O(l)
H_2C_2O_4(aq)rarr2CO_2(g)+2H^+(aq)+2e^-
Multiply each half-reaction by an integer that will allow the electrons in each half-reaction to cancel out when the half-reactions are added.
2MnO_(4)^(-)(aq)+16H^+(aq)+10e^(-)rarr2Mn^(2+)(aq)+8H_2O(l)
5H_2C_2O_4(aq)rarr10CO_2(g)+10H^+(aq)+10e^-
Finally, add the two half-reactions and simplify.
2MnO_(4)^(-)(aq)+16H^+(aq)+10e^(-)+5H_2C_2O_4(aq)rarr2Mn^(2+)(aq)+8H_2O+10CO_2(g)+10H^+(aq)+10e^-
2MnO_(4)^(-)(aq) + 6cancel(16)H^+(aq) + cancel(10e^(-)) + 5H_2C_2O_4(aq) rarr 2Mn^(2+)(aq) + 8H_2O(l) + 10CO_2(g) + cancel(10H^+(aq)) + cancel(10e^-)
2MnO_(4)^(-)(aq) + 5H_2C_2O_4(aq) + 6H^+(aq) rarr 2Mn^(2+)(aq) + 10CO_2(g) + 8H_2O(l)