Verified answer

3x² + 2xy + y² = 2

if x=1

3+2y+y^2 = 2

y^2+2y+1=0

(y+1)^2 = 0

y=-1

3x² + 2xy + y² = 2

differentiate both sides with respect to x

6x + 2y + 2x dy/dx + 2y dy/dx = 0

dy/dx (2x+2y) = -(6x+2y)

dy/dx = (-3x-y) / (x+y)

Substitute x=1 and y=-1

dy/dx = (-3+1) /(1+1) = -2/2 = -1

Replace x with 1 in your equation

So if you have dy/dx = y' = (-3x-y)/(x+y) then,

y'(1) = (-y-3)/(y+1)