Question 1027550
Differentiate once,
1.{{{4x^3+4y^3y[1]=0}}}
Differentiate twice,
{{{12x^2+4(y^3y[2]+y[1]*3y^2*[1])=0}}}
where {{{y[1]}}} is the first derivative, {{{y[2]}}} is the second.
{{{y^3y[2]=-3(y^2(y[1])^2+x^2))}}}
{{{y[2]=(-3(y^2(y[1])^2+x^2))/y^3)}}}
Then from (1),
{{{y[1]=-x^3/y^3}}}
{{{y[1]^2=x^6/y^6}}}
Substituting,
{{{y[2]=(-3(y^2(x^6/y^6)+x^2))/y^3)}}}
{{{y[2]=(-3((x^6/y^4)+x^2))/y^3)}}}
{{{y[2]=(-3x^2((x^4/y^4)+1))/y^3)}}}
{{{y[2]=(-3x^2(x^4+y^4))/y^7)}}}