At what point is the tangent to the curve x+y^2=1 parallel to the line x+2y=0?

Question

x+y^2=1
1+2y(y')=0
y'=-1/(2y)

I found the derivative so what do I do next?

Answer 1

Derivating
1+2yy´=0
so y´=-1/2y This must be equal the slope of the given line which is -1/2
so -1/(2y)=-1/2 and y =1
going back to the equation of the curve
x+1=1 so x=0
The point is (0,1)

Answer 2

the slope of the line -1/2
So this is the slope of the tangent line
1+2y(y')=0
y'=-1/(2y)=-1/2

y=1

x+1^2=1
x=0

point (0,1)