The factor of 9 in the denominator then implies 14y^2 is a multiple of 9=3^2 since 9x^2 is clearly a multiple of 9. 14 is coprime to 9 so then y^2 is a multiple of 9, or y is a multiple of 3.  gcd(x,z)=1 follows from gcd(x,y)=1 being given.

Let z=3y, then (9x^2 + 14y^2)/(9*xy) = (x^2 + 14z^2)/(xz).  

(x^2 + 14z^2)/x = x + 14*z^2/x must be an integer multiple of (x^2 + 14z^2)/(xz).  However since x is coprime to z, then x must be a factor of 14.

Similarly, (x^2 + 14z^2)/z = 1*x^2/z + 14z must also be n integer multiple of (x^2 + 14z^2)/(xz).  However since x is coprime to z, then z must be a factor of 1. And then y is 3 times a factor of 1.

Then there are only four cases to try: (x,y)=(1,3), (2,3), (7,3), (14,3).

(1,3): x/y + 14y/(9x) = 1/3 + 42/9 = 5.

(2,3): x/y + 14y/(9x) = 2/3 + 42/18 = 3.

(7,3): x/y + 14y/(9x) = 7/3 + 42/63 = 3.

(14,3): x/y + 14y/(9x) = 14/3 + 42/126 = 5.

Thus, all possible pairs (x,y) sought in the problem are (x,y)=(1,3), (2,3), (7,3), (14,3).