I need help on this calculus exercise

Find the particular solution of the differential equation that satisfies the given initial condition.

dy/dx = 2xy^2, y = 1 when x = 2

dy/dx = 2xy^2
dy/y^2 = 2x * dx

Integrate

-1 / y = x^2 + C
y = -1 / (x^2 + C)

y = 1
x = 2

1 = -1 / (2^2 + C)
1 = -1 / (4 + C)
4 + C = -1
C = -5

y = -1 / (x^2 - 5)

try 2xdx=dy/y^2