The fastest way to do this is probably plugging in numbers. Let y=8, x=4, z=2. All of the choices will work out to integers except for B which will give you (8+2)/4=5/2. This will work as long as you are not unfortunate enough to choose numbers where x=z. A more mathematical argument follows:
If x is a factor of y, then y is a multiple of x. Also, x is a multiple of z. So, y=nx for some positive integer n, and x=zm for some positive integer m. Finally because y is a multiple of x and x is a multiple of z, then y must be a multiple of z, so y=zj for some positive integer j, specifically j=mn.
A.(x+z)/z = (zm+z)/z = m+1 which is always an integer. In words, the sum of two multiples of z is always divisible by z.
B. (y+z)/x = (nx+z)/x = nx/x+z/x = n+z/x =n+1/m. This would be an integer only if m=1. If we are trying to divide a sum by x, and one of the things we are adding is divisible by x, the whole sum will be divisible by x if and only if the other thing is ALSO divisible by x. We know that y is divisible by x because y=nx. However, x is a multiple of z, so the only way z could be divisible by x is if z=x. This is possible but certainly not necessarily true.
C. (x+y)/z = (zm+zj)/z = m+j which is an integer. Again, the sum of two multiples of z is always divisible by z.
D. xy/z = (zm*zj)/z = zmj which is always an integer. In words, the product of a multiple of z and any other integer will always be divisible by z.
E. yz/x= nx*z/x = nz which is always an integer. In words, the product of a multiple of x and any other integer will always be divisible by x.
