Yes

Should be dy/dx, not dydx.

So I'm not sure if right side is (y−1)(y+1) or (y−1)/(y+1) since you already forgot one division symbol in dy/dx.

dy/dx = (y−1)(y+1)

1/((y−1)(y+1)) dy = dx

1/2 (1/(y−1) − 1/(y+1)) dy = dx

(1/(y−1) − 1/(y+1)) dy = 2 dx

∫ (1/(y−1) − 1/(y+1)) dy = ∫ 2 dx

ln|y−1| − ln|y+1| = 2x + c

ln|(y−1)/(y+1)| = 2x + c

(y−1)/(y+1) = Ce^(2x)

y(1) = 0

−1/1 = Ce²

C = −1/e²

(y−1)/(y+1) = −e^(2x)/e²

ye² − e² = −ye^(2x) − e^(2x)

ye^(2x) + ye² = e² − e^(2x)

y(e^(2x) + e²) = e² − e^(2x)

y = (e² − e^(2x)) / (e^(2x) + e²)

K