If (x^2)+2xy+(3y^2)=2, find y' and y'' when y=1?

x² + 2xy + 3y² = 2
when y = 1, 
x² + 2x*1 + 3*1² = 2
x² + 2x + 1 = 0
x = -1

x² + 2xy + 3y² = 2
2x + 2( xy’ + y ) + 6y y’ = 0
2x + 2xy’ + 2y + 6yy’ = 0

when y = 1, x = -1
2(-1) + 2( -y’ + 1 ) + 6y’ = 0
-2 - 2y’ + 2 + 6y’ = 0
y’ = 0
━━━

2x + 2xy’ + 2y + 6yy’ = 0
2 + 2( xy’’ + y’ ) + 2y’ + 6( yy’’ + y’y’) = 0
2 + 2xy’’ + 2y’ + 2y’ + 6yy’’ + 6y’y’ = 0
2 + 2xy’’ + 4y’ + 6yy’’ + 6y’y’ = 0

when y = 1, x = -1, y’ = 0
2 + 2(-1)y’’ + 4(0) + 6(1)y’’ + 6(0)(0) = 0
y’’ = -½
━━━━

Why y = 1:
x² + 2x(1) + 3(1)² = 2
x² + 2x + 1 = 0
(x + 1)² = 0
x = -1

x² + 2xy + 3y² = 2
(d/dx)(x² + 2xy + 3y²) = (d/dx)2
2x + (2xy' + 2y) + 6yy' = 0
2xy' + 6yy' = -2x - 2y
y' = -(x + y) / (x + 3y)

When y = 1, and x = -1:
y' = -[(-1) + (1)] / [(-1) + 3(1)]
y' = 0

y" = (d/dx)[-(x + y) / (x + 3y)]
y" = -[(x + 3y)(x + y)' - (x + y)(x + 3y)' / (x + 3y)²]
y" = -[(x + 3y)(1 + y') - (x + y)(1 + 3y') / (x + 3y)²]

When y = 1, x = -11 and y' = 0:
y" = -{[(-1) + 3(1)][1 + (0)] - [(-1) + (1)][1 + 3(0)] / [(-1) + 3(1)]²}
y" = -(2 - 0) / 4
y" = -1/2

Hence, when y = 1, y' = 0 and y" = -1/2

x^2 +2xy +3y^2 = 2...(1). Notation : ID = ''implicit differentiation of''.;
ID(1) gives 2x + 2y + 2xy' + 6yy' = 0, ie., (x+3y)y' +(x+y) = 0...(2).;
ID(2) gives (x+3y)y'' + (1+3y')y' + 1+ y' = 0, ie., (x+3y)y'' + (2+3y')y' +1 = 0...(3).;
When y = 1, (1) becomes x^2 +2x +1 = 0, ie., (x+1)^2 = 0 and x = -1.;
For (x,y) = (-1,1), (2) becomes (-1+3)y' +(-1+1) = 0, ie., 2y' = 0, ie., y' = 0.;
For (x,y) = (-1,1), (3) becomes (-1+3)y'' + 1 = 0, ie., 2y'' = -1, ie., y'' = -(1/2).

Take derivatives with respect to x on both sides:

    2x + (2y + 2xy') + 6yy' = 0
    (2x + 6y) y' = -2x - 2y
    y' = - (x + y) / (x + 3y)

The value of x matters, so solve for what it must be if y = 1:

    x^2 + 2x + 3 = 2    .... plug in y=1
    x^2 + 2x + 1 = 0
    (x + 1)^2 = 0
    x = -1

So yes, y' = 0 because (x + y) in the numerator is (-1 + 1) = 0.  Take derivatives with respect to x again in that first equation above:

    2 + 2y' + (2xy'' + 2y') + 6[(y')^2 + yy''] = 0
    2 + 0  - 2y'' + 0 + 6(0^2 + y'') = 0
     2 - 2y'' + 6y'' = 0
    4y'' = -2
    y'' = -1/2