How would you find the net ionic equation of "HCl" + "Zn"

For starters, make sure that you have a balanced equation to work with. To balance the equation given to you, multiply the hydrochloric acid by #2#

#"Zn"_ ((s)) + color(blue)(2)"HCl"_ ((aq)) -> "ZnCl"_ (2(aq)) + "H"_ (2(g))#

You know that hydrochloric acid is a strong acid, which means that it dissociates completely in aqueous solution to produce hydrogen ions, #"H"^(+)#, and chloride anions

#"HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#

Now ,zinc chloride, #"ZnCl"_2#, is soluble in aqueous solution, which means that it too will exist as ions

#"ZnCl"_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#

This means that you can write

#"Zn"_ ((s)) + color(blue)(2) xx ["H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)] -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ (2(g))#

This is equivalent to

#"Zn"_ ((s)) + 2"H"_ ((aq))^(+) + 2"Cl"_ ((aq))^(-) -> "Zn"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ (2(g))#

Now, the net ionic equation is obtained by removing the spectator ions, i.e. the ions that are present on both sides of the equation.

In this case, you have

#"Zn"_ ((s)) + 2"H"_ ((aq))^(+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) -> "Zn"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) + "H"_ (2(g))#

The net ionic equation that describes this single replacement reaction will thus be

#color(darkgreen)(ul(color(black)("Zn"_ ((s)) + 2"H"_ ((aq))^(+) -> "Zn"_ ((aq))^(2+) + "H"_ (2(g)))))#