What is the solution to this initial value problem?

You don't need to solve the equation at all.

Note y≡1 (constant function) satisfies both the equation and initial condition:

(1)' = 0 = 1²-1,

y(0) = 1.

The solution y=1 is the only solution for the initial problem since it easy to check that both functions

F(x,y)=y²-1,

∂F(x,y)/∂y = 2y are continuous in R²

:)

dy/dx = y^2 -1
separate the variables
dy/(y^2-1) = dx
dy/ ((y-1)(y+1)) = dx

∫ dy/ ((y-1)(y+1)) = ∫ dx

Apply partial fractions to integrate the left side
1 /((y-1)(y+1)) = A/(y-1) + B/(y+1)
multiply both sides by (y-1)(y+1)
1 = A(y+1) + B(y-1)

Let y=-1
1= A(0) + B(-2)
1= -2B
B = -1/2

Let y=1
1 = A(2) + B(0)
2A = 1
A = 1/2

1 /((y-1)(y+1)) = A/(y-1) + B/(y+1)
∫ dy/ ((y-1)(y+1)) = (1/2) ∫ dy/(y-1) - (1/2) ∫ dy/(y+1)
= (1/2) ln (y-1) - (1/2) ln (y+1)
= (1/2) ln ( (y-1)/(y+1) )

∫ dy/ ((y-1)(y+1)) = ∫ dx
(1/2) ln ( (y-1)/(y+1) ) = x + C
ln ( ((y-1)/(y+1))^(1/2) ) = x + C
((y-1)/(y+1) )^(1/2) = e^x e^C
((y-1)/(y+1) )^(1/2) = C1 e^x
(y-1)/(y+1) = C2 e^(2x)
(y-1) = C2 e^(2x) (y+1)
y ( 1 - C2 e^(2x) ) = C2 e^(2x) + 1
y = (C2 e^(2x) + 1) / ( 1 - C2 e^(2x) )

y(0) = 1
1 = (C2 +1)/(1-C2)
C2 = 0
cannot proceed further