#y = abs(x^2-1) = {(x^2-1," if ",x^2-1 > 0),(1-x^2," if ", x^2-1 < 0):}#

Solving the inequality #x^2-1 > 0# gets us,

#y = {(x^2-1," if ",x < -1),(1-x^2," if ",-1 < x < 1),(x^2-1," if ",x > 1) :}#

So the derivative is given by

#y' = {(2x," if ",x < -1),(-2x," if ",-1 < x < 1),(2x," if ",x > 1) :}#

#2x=1# at #x=1/2# but the derivative at #x=1/2# is not given by #2x#.

#-2x = 1# at #x= -1/2# and at #x=-1/2#, the derivative **is** given by #-2x#.

So the slope is #1# and #x=-1/2#. At that point, we get #y = 3/4#.

So the only point on the curve at which the slope is #1# is #(-1/2,3/4)#.