First, rewrite the expression as shown below and simplify the constants:

#(15/40)(x^4/x^3)(y^2/y^3) = (5 xx 3)/(5 xx 8)(x^4/x^3)(y^2/y^3) =#

#(color(red)(cancel(color(black)(5))) xx 3)/(color(red)(cancel(color(black)(5))) xx 8)(x^4/x^3)(y^2/y^3) = 3/8(x^4/x^3)(y^2/y^3)#

Now, use these two rules of exponents to simplify the #x# and #y# terms:

#x^color(red)(a)/x^color(blue)(b) = x^(color(red)(a)-color(blue)(b))# and #x^color(red)(a)/x^color(blue)(b) = 1/x^(color(blue)(b)-color(red)(a))#

#3/8(x^color(red)(4)/x^color(blue)(3))(y^color(red)(2)/y^color(blue)(3)) = 3/8(x^(color(red)(4)-color(blue)(3)))(1/x^(color(blue)(3)-color(red)(2))) = 3/8(x^1)(1/y^1) = #

#(3x^1)/(9y^1)#

Now, we can use this rule of exponents to complete the simplification:

#a^color(red)(1) = a#

#(3x^color(red)(1))/(9y^color(red)(1)) = (3x)/(8y)#

Because we cannot divide by #0#, from the original expression #40x^3y^3 != 0#, therefore #x != 0# and #y != 0#