The correct options are
B xyx+y2
C x−ycosxysinx+y
Consider f(x) to be homogeneous.
Then it should satisfy the condition, f(vx)=vf(x) or in other words, replacing x by (vx) and y by (vy) should give us v times the same initial expression, where v is an arbitrary constant.
For example consider option A
Replacing x by (vx) and y by (vy)
vx+vycos(vyvx)
=v(x+ycos(yx))
The degree of all the terms is same.
However these conditions are not met by B and C.
Hence Options B and C are non-homogenous.