Verified answer

Including parentheses would really help me figure out if you have three separate fractions or one fraction with a fraction in both the numerator and the denominator. I'm going to assume that you know that 3/1 = 3 and solve the version with a fraction in both numerator and denominator.

(3xy + y)/((x - y) - 3) / (1 - (x - 3y))/(x + y) =

[(3xy + y)/((x - y) - 3)] / [(1 - (x - 3y))/(x + y)] =

[(3xy + y)/(x - y - 3)] / [(1 - x + 3y)/(x + y)] =

[(3xy + y)/(x - y - 3)] * [(x + y)/(1 - x + 3y)] =

[(3xy + y)/(x - y - 3)] * [(x + y)/(1 - x + 3y)] =

[(3xy + y)(x + y)] / [(x - y - 3)(1 - x + 3y)] =

[(3xy)(x) + (3xy)(y) + (y)(x) + (y)(y))] / [(x - y - 3)(1 - x + 3y)] =

[3x²y + 3xy² + xy + y²] / [x - x² - 3xy - y + xy - 3y² - 3 - 3x - 9y] =

[3x²y + 3xy² + xy + y²] / [x - x² - 2xy - y - 3y² - 3 - 3x - 9y] =

[3x²y + 3xy² + xy + y²] / [-x² - 2xy - y - 3y² - 3 - 2x - 9y] =

[3x²y + 3xy² + xy + y²] / [-x² - 2xy - 3y² - 3 - 2x - 10y] =

[3x²y + 3xy² + xy + y²] / [-x² - 3y² - 2xy - 2x - 10y - 3]

Hope this helps you!