Here's what I got.
In this particular case, the complete ionic equation and the net ionic equation are the same because there are no spectator ions.
So for this chemical equation
#"Ba"("OH")_ (2(aq)) + "H"_ 2"SO"_ (4(aq)) -> "BaSO"_ (4(s)) darr + 2"H"_ 2"O"_ ((l))#
you will have
#"Ba"_ ((aq))^(2+) + 2"OH"_ ((aq))^(-) + 2"H"_ ((aq))^(+) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr + 2"H"_ 2"O"_ ((l))#
As you can see, this reaction has no spectator ions because none of the ions are present on both sides of the equation, which is why you can say that the complete ionic equation and the net ionic equation are the same.
So if you can't break up covalent compounds or precipitates, just focus on the ionic compounds that you can break up and write the complete ionic equation using all the ions present in the solution and the net ionic equation by removing the spectator ions, i.e. the ions present on both sides of the equation.
It's worth mentioning here that barium hydroxide, which is a strong base, it actually not that soluble in water. The amount of barium hydroxide that does dissolve will dissociate completely, so you can say that you have
#"Ba"("OH")_ (2(s)) rightleftharpoons "Ba"_ ((aq))^(2+) + 2"OH"_ ((aq))^(-)#
Now, barium hydroxide is soluble in acid because of Le Chatelier's Principle. When you add the sulfuric acid, the hydrogen ions produced by the acid will react with the hydroxide anions produced by the dissociation of the base to produce water.
This will cause the equilibrium to shift to the right, i.e. more barium hydroxide will dissolve to produce barium cations and hydroxide anions.