Which type of reaction is #AlCl_3 + 3 KOH ->Al(OH)_3+ 3 KCl#?

none of the atoms or polyatomic ions change the oxidations states, so it is not a oxidation reduction reaction.

The reaction follows the general form of a double replacement

# A^+B^- (+) C^+ D^-#= # A^+ D^- (+ ) C^+ B^-#
The positive three ion of Al disassociates with the Chlorine -1 ion.
then the still positive +3 ion of Al then combines with the -1 OH

The positive +1 K disassociates with the Hydroxide ion -1 ion then the still positive +1 K ion combines with the -1 Chlorine ion.

This is a classic example of a double replacement reaction.

In a double replacement reaction, the cations in two compounds switch places. The generic reaction is:

#color(red)"A"color(green)"X" + color(blue)"B"color(purple)"Y""##rarr##color(blue)"B"color(green)"X" + color(red)"A"color(purple)"Y"#,

where #color(red)"A"# and #color(blue)"B"# are the cations and #color(green)"X"# and #color(purple)"Y"# are the anions.

One of the products of a double replacement reaction must be water (neutralization), an insoluble gas (such as #"H"_2"# or #"CO"_2"#, or a precipitate (an insoluble solid).

#color(red)"Al"color(green)("Cl")_3("aq") + 3color(blue)"K"color(purple)("OH")("aq)"##rarr##"3"color(blue)("K")color(green)("Cl") + color(red)("Al")(color(purple)("OH"))"_3#

When the aqueous #"AlCl"_3"# and aqueous #"KOH"# solutions are mixed, an insoluble solid is produced. We can see that neither of the products is water or an insoluble gas, but one of the products is a precipitate because it is insoluble in aqueous solutions.

We can find out which product is the precipitate by consulting a solubility table like the one below. We can see that compounds containing alkali metal cations, such as #"K"^(+)"#, are soluble in aqueous solutions. We can also see that compounds that contain the hydroxide ion, #"OH"^(-)"#, are insoluble, and aluminum is not one of the exceptions. Therefore, the precipitate is the compound #"Al(OH)"_3"#.

The entire equation with physical states included is:

#"AlCl"_3("aq") + "3KOH(aq)"##rarr##"Al(OH)"_3("s") + "3KCl(aq)"#

We can use a down arrow to show that the solid is a precipitate.

#"AlCl"_3("aq") + "3KOH(aq)"##rarr##"Al(OH)"_3darr + "3KCl(aq)"#