<s>[INST] Lost in second derivative simplification My book has the following problem:
Find the second derivative of y for $x^2 + xy + y^2 = 3$
I follow them this far:
$y'' = -\dfrac{ 6(x^2 + xy + y^2) }{ x + 2y }$
But then how do they get here?
$y'' = -\dfrac{ 18 }{ (x + 2y)^3}$

Updated
Assuming the $x^2 + xy + y^2$ is substituted in, then shouldn't it read
$y'' = -\dfrac{ 18 } { x + 2y } = -(\dfrac{ 3 }{ (x + 2y)})^3$
or something like that? I don't see how the bottom gets cubed but the top stays 18
 [/INST] I must have a different method than your book since I never reached the same step as you, but here is my derivation if it helps:
$x^2 + xy + y^2 = 3$
Find $\dfrac{dy}{dx}$:
$2x + [(x)(y') + (y)(1)] + 2y(y') = 0$
$2x + xy' + y + 2yy' = 0$
$(2x + y) + y'(x + 2y) = 0$
Solve for $y'$:
$y' = -\dfrac{ 2x + y }{ x + 2y }$
Find $y''$:
$y'' = -\dfrac{ (x + 2y)(2 + y') - (2x + y)(1 + 2y') }{ (x + 2y)^2 }$
$y'' = -\dfrac{ (2x + xy' + 4y + 2yy') - (2x + 4xy' + y + 2yy') }{ (x + 2y)^2 }$
$y'' = -\dfrac{ 2x + xy' + 4y + 2yy' - 2x - 4xy' - y - 2yy' }{ (x + 2y)^2 }$
$y'' = -\dfrac{ -3xy' + 3y }{ (x + 2y)^2 }$
Substitute for $y'$:
$y'' = -\dfrac{ -3x(-\dfrac{ 2x + y }{ x + 2y }) + 3y }{ (x + 2y)^2 }$
$y'' = -\dfrac{ \dfrac{ 6x^2 + 3xy }{ x + 2y } + 3y }{ (x + 2y)^2 }$
$y'' = -\dfrac{ \dfrac{ 6x^2 + 3xy + 3y(x + 2y) }{ x + 2y } }{ (x + 2y)^2 }$
$y'' = -(\dfrac{ 6x^2 + 3xy + 3xy + 6y^2 }{ x + 2y })(\dfrac{ 1 }{ (x + 2y)^2 })$
$y'' = -\dfrac{ 6x^2 + 6xy + 6y^2 }{ (x + 2y)^3 }$
$y'' = -\dfrac{ 6(x^2 + xy + y^2) }{ (x + 2y)^3 }$
As other users noted, replace $x^2 + xy + y^2 = 3$ from the original equation:
$y'' = -\dfrac{ 6(3) }{ (x + 2y)^3 }$
$y'' = -\dfrac{ 18 }{ (x + 2y)^3 }$
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