Rewriting using standard symbols:

#color(white)("XXXX")##(-105 div 35 + 6)(9)#

Everything inside the parentheses is done first with parentheses evaluated left to right

So #(-105 div 35 + 6)# is evaluated first

#color(white)("XXXX")#Within the parentheses multiplication and division is done before addition and subtraction

#color(white)("XXXX")#So #-105 div 35# is done first

#color(white)("XXXX")##color(white)("XXXX")##-105 div 35 = -3#

#color(white)("XXXX")##(-105div 35 + 6)# therefore becomes #(-3 +6)#

#color(white)("XXXX")##color(white)("XXXX")##-3+6 = 3#

#color(white)("XXXX")##(-105 div 35 +6)# becomes #(-3+6)# becomes #(3)# or just #3#

Moving to the second set of parentheses:

we have nothing to evaluate except #(9) = 9#

We are left with an implied multiplication

#color(white)("XXXX")##3xx9 = 27#