implicit differention?

Question

Find dy/dx by implicit differentiation.
1+x=sin(xy^2)

i worked it out and got as far as
1=cos(xy^2)(y^2+2xy)(dy/dx)....i tried to simplify it and got
2xy((1/cosxy^2)-y^2) but i got it wrong

Answer 1

1= cos(xy^2) *(y^2+2xy*y´)so y´=[1-y^2*cos(xy^2)]/[2xy*cos(xy^2)]

Answer 2

1+x = sin(xy^2)

differentiating implicitly

1 = cos(xy^2)*(x* 2yy' + y^2)

1 = 2xy y' cos(xy^2) + y^2 cos(xy^2)

2xy y' cos(xy^2) = 1 - y^2 cos(xy^2)

y' = [1 - y^2 cos(xy^2)]/2xy cos(xy^2)

y' = (1/2xy) sec(xy^2) - (y/2x)