It is given that

Now, f’(x) =0

The points x =1 and x = -1 divide the real line in three disjoint intervals

(-∞,-1),(-1,1) and (1,∞)

Now in interval, (-1,1)

it is clear that -1 < x < 1

⇒ x² < 1

Therefore, f’(x) = 1- < 0 (-1,1) ~ {0}

Therefore, f is strictly decreasing on (-1,1) ~ {0}

x < -1 or 1 < x

⇒ x² > 1

Therefore, f’(x) = 1- > 0 (-∞, -1) and (1,∞)

Therefore, f is strictly increasing in interval I disjoint from (-1,1)

Hence Proved.