You need to bring the fractions to the right to a common denominator, such that:

`1/y - 1/(y+1) = (y + 1)*1/(y*(y + 1)) - y*1/(y*(y + 1))`

`1/y - 1/(y+1) = (y + 1 - y)/(y*(y + 1))`

Reducing duplicate terms yields:

`1/y - 1/(y+1) = 1/(y*(y + 1))`

**Hence, testing if `1/(y*(y + 1)) = 1/y - 1/(y+1)` yields that the statement is valid.**

We'll solve the problem starting from the idea that the fraction 1/y(y+1) is the result of addition or subtraction of 2 elementary fractions:

1/y(y+1) = A/y + B/(y+1) (1)

We'll multiply the ratio A/y by (y+1) and we'll multiply the ratio B/(y+1) by y.

1/y(y+1) = [A(y+1) + By]/y(y+1)

Since the denominators of both sides are matching, we'll write the numerators, only.

1 = A(y+1) + By

We'll remove the brackets:

1 = Ay + A + By

We'll factorize by a to the right side:

1 = y(A+B) + A

If the expressions from both sides are equivalent, the correspondent coefficients are equal.

A+B = 0

A = 1

1 + B = 0

B = -1

We'll substitute A and B into the expression (1):

1/y(y+1) = 1/y - 1/(y+1)

**Since the LHS=RHS, the identity was verified.**