Precalculus, Chapter 6, 6.3, Section 6.3, Problem 41

Hence, you need to find the unit vector having the same direction as the vector v = <-2,2> , hence, you need to use the formula, such that:
u = v/|v|
You need to evaluate the magnitude |v|, such that:
|v| = sqrt(a^2+b^2)
|v| = sqrt((-2)^2 + 2^2) => |v| = sqrt(4+4) => |v| = sqrt 8 => |v| = 2sqrt2
u = (<-2,2>)/(2sqrt2) => u = <-2/(2sqrt2), 2/(2sqrt2)>
u = <-1/(sqrt2), 1/(sqrt2)>
You need to check that the magnitude of the unit vector is 1, such that:
|u| = sqrt((-1/(sqrt2))^2 + (1/(sqrt2))^2)
|u| = sqrt(1/2 + 1/2)
|u| = sqrt (1)
|u| = 1
Hence, evaluated the unit vector yields u = <-1/(sqrt2), 1/(sqrt2)>.