My MS SQL solution:

SELECT f1.X, f1.Y FROM Functions f1
INNER JOIN Functions f2 ON f1.X=f2.Y AND f1.Y=f2.X
GROUP BY f1.X, f1.Y
HAVING COUNT(f1.X)>1 or f1.X<f1.Y
ORDER BY f1.X 

The question should be made a little more clear. When selecting the symmetric pair, it assumes x<=y . This is not mentioned in the problem. (for the pair x1,y1 and y1,x1 select only the one which has x<=y) Also for y=f(x), if x=y, it can be assumed that f(y)=x, however we need to find two pairs of x,y to consider that. This works as a great SQL problem, but the scenario is incorrect. We cannot think of this as a mathematical function.

Here's a simple solution that avoids using joins or subqueries:

SELECT A.x, A.y
FROM FUNCTIONS A JOIN FUNCTIONS B ON
    A.x = B.y AND A.y = B.x
GROUP BY A.x, A.y
HAVING COUNT(A.x) > 1 OR A.x < A.y
ORDER BY A.x

The key is the HAVING line, with two conditions.

1. The first condition in that line makes sure pairs with the same X and Y values don't get to count themselves as their symmetric pair. (e.g. if 10 10 appears one time it's not counted as a symmetric pair, but as 13 13 appears twice, that is a symmetric pair)

2. The second condition ensures that only one of a pair is displayed. (e.g. if 3 24 and 24 3 form a symmetric pair, it will only display 3 24, not 24 3 as well.)

My solution:

select x, y from functions f1 
    where exists(select * from functions f2 where f2.y=f1.x 
    and f2.x=f1.y and f2.x>f1.x) and (x!=y) 
union 
select x, y from functions f1 where x=y and 
    ((select count(*) from functions where x=f1.x and y=f1.x)>1)    
        order by x;

My simple solution using SQL Server:

;WITH CTE_Functions
AS
(
	SELECT X, Y, ROW_NUMBER() OVER (PARTITION BY (X + Y), ABS(X-Y) ORDER BY X DESC) AS RowNumber
	FROM Functions WITH (NOLOCK)	
)
SELECT DISTINCT X, Y
FROM CTE_Functions
WHERE RowNumber > 1
AND X <= Y

This solution handles all possibilities including duplicates and negative values. No need to use JOINs or Sub queries or any complex logic. It is very straightforward, concise and simple solution.