Since a+b+c=0, c=-a-b. Substituting for c in the first two terms: a²/(2a²-ab-b²)+b²/(2b²-ab-a²).
The symmetry in the denominators suggest that a-b is a factor (because putting a=b reduces the denominator to zero) in each case, so we can divide by this factor to find the other factor:
a²/((a-b)(2a+b))-b²/((a-b)(a+2b)).
So we have:
(a²(a+2b)-b²(2a+b))/((a-b)(2a+b)(a+2b))=
(a³+2a²b-2ab²-b³)/((a-b)(2a+b)(a+2b)). This time the numerator factorises as a-b is a factor because of symmetry:
(a-b)(a²+3ab+b²), so (a-b) is a common factor in this fraction which now reduces to:
(a²+3ab+b²)/((2a+b)(a+2b)).
The numerator is the same as (a+b)²+ab=c²+ab. Also, (2a+b)(a+2b)=(a-c)(b-c) when we substitute a+b=-c. Now we can add this result to the third term: (c²+ab)/((a-c)(b-c))+c²/(2c²+ab).
Expand (a-c)(b-c)=ab-c(a+b)+c²=ab+c²+c²=ab+2c².
So we have: (c²+ab)/(ab+2c²)+c²/(2c²+ab)=(2c²+ab)/(2c²+ab)=1 QED.
Note that if one of a, b or c is zero in the original expression the result is also 1 (½+½=1).