find x^2-y^2/x-y at x=y?

(x² - y²)/ x - y at x = y

(x² - y²) = (x - y) (x + y)

[(x - y) (x + y)]/ (x - y) = x + y

put x = y

x + x = 2x

or

y + y = 2y

Factor and cancel down

(x - y)(/x + y) / (x - y) =>

(x + y)

At x = y

Then substitute

(y + y) = 2y

Assuming the expression is (x² - y²)/(x-y) (per your answer to your own question)

At x = y, the expression is undefined because x-y = 0 and division by zero is an illegal operation. However, the expression does have a limit as x approaches y:

lim(x→y) (x² - y²)/(x-y)

= lim(x→y) (x-y)(x+y)/(x-y)

= lim(x→y) (x+y)

= 2y

(x^2 - y^2)/(x - y)

= (x + y)

When x = y you get

x^2-y^2/x-y =

y^2 - y^2/y - y =

y^2 - y - y = (for y not equal to zero)

y^2 - 2y for y not equal to zero. It is undefined when x = y = 0

If you meant (x^2-y^2)/(x-y) then this is undefined when y = x as you would get division by zero.

factorise numerator leads to (x+y)(x-y)/(x-y)=x+y=2x or 2y