The question appears ambiguous. I'll assume initially: (3/x)+5y=2, 7x+(6/y)=3.
3+5xy=2x, 7xy+6=3y,
21+35xy=14x, 35xy+30=15y,
Therefore: 14x-21+30=15y, 14x-15y+9=0, y=(14x+9)/15.
Substitute for y in the first equation:
3+5x(14x+9)/15=2x,
3+x(14x+9)/3=2x,
9+14x2+9x-6x=0,
14x2+3x+9=0 has no real solutions, so we need to revise the original assumption.
Let's now assume: 3/(x+5y)=2, (7x+6)/y=3.
2x+10y=3, 7x-3y=-6,
6x+30y=9, 70x-30y=-60,
76x=-51, x=-51/76, 10y=3-2x=3+51/38=165/38, y=165/380=33/76.
So the solution (if the assumption is correct) is x=-51/76, y=33/76.