Hard calculus question on derivatives

Please, I need a full solution to this:

When x^2 + xy + y^2 = a^2,
find the value of d^2y/dx^2.

x^2 + xy + y^2 = a^2

I'm assuming a is a constant and x/y are variables.

d/dx (x^2 + xy + y^2) = d/dx (a^2)
d/dx (x^2) + d/dx (xy) + d/dx (y^2) = 0
Using the Chain Rule and the Product Rule:
2x + (x'y + y'x) + 2y*y' = 0
2x + y + y'x + y'2y = 0
y'(x) + y'(2y) = -2x - y
y'(x + 2y) = -2x - y

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

Now.. do it over again:
y'(x + 2y) = -2x - y
y''(x + 2y) + y'(1 + 2y') = - 2 - y'
y''(x + 2y) + y' + 2(y')^2 = -2 - y'

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

y''(x + 2y) - (2x + y)/(2y + x) + 2[ (- 2x - y)^2 ]/[ (x + 2y)^2] = - 2 + (2x + y)/(2y + x)

After doing a ton of simplifying, I ended up with:
d2y/dx2 = ( -x(5x + y) )/( (x + 2y)^3 )

Use implicit derivatives with respect to dx to get dy/dx.
x^2 + xy + y^2 = a^2
2x + 1(y) + (dy/dx)x + 2y(dy/dx) = 0
All terms with dy/dx move to the left and all other terms move to the right. Rmember to switch signs when you switch sides. Factor out dy/dx on the left.
(dy/dx)(x + 2y) = -2x - y
Divide both sides by x+2y.
dy/dx = (-2x - y) / (x+ 2y)
Here is the first derivative. Use quotient rule to get second derivative.
d^2y / dx^2 = [ (-2 - dy/dx) (x+2y) - (1+ 2dy/dx)(-2x - y) ] / (x+2y)^2
Factor out the negative in the second term. Cancel it with the subtraction.
d^2y / dx^2 = [ (-2 - dy/dx) (x+2y) + (1+ 2dy/dx)(2x + y) ] / (x+2y)^2
Expand and simplify.
d^2y / dx^2 = [ -2x - 4y - (dy/dx)x - 2y(dy/dx) + 2x + y + 2y(dy/dx)+4x(dy/dx) ] / (x+2y)^2
d^2y / dx^2 = [ 3x(dy/dx) - 3y ] / (x+2y)^2
d^2y / dx^2 = 3[ x(dy/dx) - y ] / (x+2y)^2
Second derivative.

Differentiate implicitly as follows:

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

Differentiate again:

2 + xy'' + y' + y' + 2y'' + y'(2y') = 0

2 + xy'' + 2y' + 2y'' + 2(y')^2 = 0

xy'' + 2y'' = -2 - 2y' - 2(y')^2

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

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

Use Implicit rule
2 + xd^2y+dy+ dy + 2d^2y + dy(2dy) = 0

2 + xd^2y+ 2dy + 2d^2y + 2(dy)^2 = 0

xd^2y + 2d^2y= -2 - 2dy - 2(dy)^2

d^2y(x + 2) = -2(1 + dy + (dy)^2)

d^2y = -2(1 + dy + (dy)^2) / (x + 2)