Calculus question......please help..........10 pts

if x^2 + xy + 3y^2 =1 prove that (x+6y)^3 y" +22 = 0 , here y" means double differentiation of y w.r.t. x
please give the steps too.

x² + xy + 3y² = 1

Differentiate.

2x + y + xy' + 6yy' = 0
(x + 6y)y' = -(2x + y)
y' = -(2x + y)/(x + 6y)
Set that aside for now.

2x + y + xy' + 6yy' = 0

Differentiate.

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

Substitute y' = -(2x + y)/(x + 6y).

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

Multiply by (x + 6y)², expand, and group.

(x + 6y)³y'' - 2(2x + y)(x + 6y) + 6(2x + y)² + 2(x + 6y)² = 0
(x + 6y)³y'' - 4x² - 26xy - 12y² + 24x² + 24xy + 6y² + 2x² + 24xy + 72y² = 0
(x + 6y)³y'' + 22x² + 22xy + 66y² = 0
(x + 6y)³y'' + 22(x² + xy + 3y²) = 0

Substitute x² + xy + 3y² = 1.

(x + 6y)³y'' + 22 = 0

if x^2 + xy + 3y^2 =1 prove that (x+6y)^3 y" +22 = 0 --->

(1) d(x^2 + xy + 3y^2)/dx = 2x + xy' + y + 6yy' = d(0)/dx = 0

(2) y' (x + 6y) = - (2x + y) ---> y' = - (2x + y)/(x + 6y)

(3) y" (x + 6y) + y' (1 + 6y') = - (2 + 6y')

(4) y" (x + 6y) + y' (1 + 6y') + (2 + 6y') = 0 = y" (x + 6y) + y' (1 + 6y') + (1 + 6y') + 1

(5) y" (x + 6y) + y' (1 + 6y') + (1 + 6y') + 1 = y" (x + 6y) + (y' + 1)(1 + 6y') + 1 = 0

(6) y" (x + 6y) + (y' + 1)(1 + 6y') + 1 = 0 --->

y" (x + 6y) + [- (2x + y)/(x + 6y) + 1][1 + 6{- (2x + y)/(x + 6y)] + 1 = 0

(7) y" (x + 6y)^2 + [- (2x + y)] + (x + 6y) + 6[- (2x + y)] + (x + 6y) = 0 --->

(8) y" (x + 6y)^2 - 2x - y + x + 6y - 12x - 6y + x + 6y = 0 --->

(9) y" (x + 6y)^2 - 12x + 5y = 0