Verified answer

Take the implicit derivative:

2yy' + 4xy' + 4y + 2x = 0

2yy' + 4xy' = -2x - 4y

y'(2y + 4x) = -2x - 4y

y' = (-2x - 4y)/(4x + 2y)

y' = (-x - 2y)/(2x + y)

Plug in (2, 1)

y' = (-2 - 2(1))/(2(2) + 1)

y' = -4/5

Use implicit differentiation for getting dy/dx .- Apply d/dx to both sides , take in mind that y =Y(x) , so

2y dy/dx +4 (xdy/dx+y )+2x= 0

(2y+4x) dy/dx= -(2x+-4y)

dy/dx= -(x+2y)/( y+2x) at (2,1)

dy/dx= -( 4/5) the line is y-1= -(4/5)(x-2)

y= -(4/5)x +13/5