Al2(co3)3 + 6hno3 = 2al(no3)3 +3co2 + 3h20
is the net ionic equation
6h+ + al(co3)3 = 2al3+ + 3co2 + 3h2o

The first step is, as you have done, to write a properly balanced molecular equation, showing the states of reactants and products.
2 Al(NO3)3(aq) + 3 Na2CO3(aq) = Al2(CO3)3(s) + 6 NaNO3(aq)

Next step is to break up each reactant and product into the respective dissociated ions. But remember that, solids, gases and water are not taken as dissociating. Pay attention to the number of ions in each compound
2Al 3- (aq) + 6NO3- (aq) + 6Na-(aq) + 3CO3 2- (aq) → Al2(CO3)3(s) + 6Na+(aq) + 6NO3-(aq)
That is the full ionic equation. Note that Al2(CO3)3 being a solid precipitate does not ionise.
Writing net ionic equation:
Now Look at each ion on the reactant side and see if it is repeated exactly unchanged on the product side:
2Al3-(aq) as an example, does not appear on the products side. Keep it.
6NO3- (aq) does appear exactly as 6NO3- on the products side - delete it on both sides.
Continue this for all the reactants side. You will be left with the NET IONIC EQUATION
2Al 3- (aq) + 3CO3 2- (aq) → Al2(CO3)3(s)
That is the net ionic equation.
The ions which are common to both sides are spectator ions, they have not taken part in any reaction, so they are eliminated from the net ionic equatio

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The answer is conduction