How do I find the empirical formula for cobalt oxalate hydrate with mass of oxalate and oxide produced?
I'm not exactly sure - but I guess you are trying to find the amount of water in your sample and to determine the ratio to cobalt oxalate.
Some parts of your calculation are correct, you do however forget that $\ce{CO2}$ is released. This would be my approach:
$$n_\ce{Co3O4} = \frac{\pu{115.8 mg}}{\pu{240.79 g/mol}} = \pu{0.4809 mmol}$$
This means that there was three times as much $\ce{CoC2O4}$ in your sample.
$$n_\ce{CoC2O4} = \pu{1.443 mmol}$$
That equals a weight of $\pu{212.1 mg}$. Your original sample weighed $\pu{328.3 mg}$, so there is $\pu{116.2 mg}$ left that seem to be $\ce{H2O}$ ($\pu{18.015 g/mol}$), giving you.
$$n_\ce{H2O} = \pu{6.45 mmol}$$
The ratio of cobalt oxalate to water then is $\left(\frac{6.450}{1.443} = 4.5\right)$ which implies the empirical formula is $\ce{2CoC2O4.9H2O}$