Any ideas!

37/13= 2+ 1/(x+1/(y+1/z))

=> 37/13 - 2 = 1/(x+1/(y+1/z))

=> 11/13 = 1/(x+1/(y+1/z))

inverting on both sides

=> 13/11 = x+1/(y+1/z)

NOTE: 13/11 is greater than 1 and less than 2

now x, y , z are positive integers therefore if x is a number greater than or equal to 2 then 1/(y+1/z) will be negative which is not possible as both y and z are positive integers.

Hence x = 1

=> 13/11 = 1+ 1/(y+1/z)

=> 2/11 = 1/(y+1/z)

now inverting on both sides we get, 11/2 = y + 1/z

now we know that 11/2 = 5 + 1/2

therefore it can also be written as , 5 + 1/2 = y + 1/z

now since y and z are integers, by comparing the terms on L.H.S and R.H.S we obtain y= 5 and z=2

therefore x+y+z=1+5+2

hence answer is 8

$\displaystyle \frac { 11 } { 13 } = \frac { 1 } { x + \frac { 1 } { y + \frac { 1 } { z } } } \rightarrow \frac { 1 } { \frac { 13 } { 11 } } = \frac { 1 } { x + \frac { 1 } { y + \frac { 1 } { z } } } \rightarrow \frac { 13 } { 11 } = x + \frac { 1 } { y + \frac { 1 } { z } } \rightarrow x = 1 \rightarrow \frac { 2 } { 11 } = \frac { 1 } { y + \frac { 1 } { z } } \rightarrow \frac { 1 } { \frac { 11 } { 2 } } = \frac { 1 } { y + \frac { 1 } { z } } \rightarrow \frac { 11 } { 2 } = y + \frac { 1 } { z } \rightarrow y = 5, z = 2 \Rightarrow x + y + z = 1 + 5 + 2 = \boxed { 8 }$ .