Since f=x^2-y^2+(2x^2y^2-k)/(x^2-y^2) = x^2-y^2 + (2k^2-k)/(x^2-y^2), using the arithmetic-geometric mean inequality, (a+b)/2 >= sqrt(ab) with a=x^2-y^2 and b=(2k^2-k)/(x^2-y^2) yields
f >= 2sqrt(2k^2 - k).
The equality holds when a = b -->
x = sqrt( (sqrt(2k^2-k) + sqrt(6k^2-k))/2 ).
As noted by Joel in his comment, the restriction y>1 forces k > 1.7640 (to 4 decimal places).
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