Question 374960
Use a common denominator, {{{x^2-y^2=(x+y)(x-y)}}}
{{{ 1/(x-y)-(2x)/(x^2-y^2 )+1/(x+y)=(x+y)/(x^2-y^2)-(2x)/(x^2-y^2 )+(x-y)/(x^2-y^2)}}}
{{{ 1/(x-y)-(2x)/(x^2-y^2 )+1/(x+y)=(x+y-2x+x-y)/(x^2-y^2)}}}
{{{ 1/(x-y)-(2x)/(x^2-y^2 )+1/(x+y)=(0)/(x^2-y^2)}}}
{{{highlight( 1/(x-y)-(2x)/(x^2-y^2 )+1/(x+y)=0)}}}