Let $f(x,y)=2x^3-4xy+2y^3$ knowing that $y=f(x)$, use the implicit function theorem to find $y''(1)$
So, if $x_0=1$, then $f(1,y_0)=2-4y_0+2y_0^3$ and $f(1,y_0)=0$ in 3 points. Lets take $y_0=1$.
- $f(1,1)=0$
- ${\partial f \over \partial y}(1,1)=2 \not = 0$, so the theorem can be used.
${\partial f \over \partial x}(1,1)=2$, and $$y'(x)= -{{\partial f \over \partial x} \over {\partial f \over \partial y}}(1,1)=-{2\over 2} = -1$$
So, now I want to get $y''(1)$. Taking $y=f(x)$ I have to apply the chain rule every time I found $y$: $$y''(x)={{(12x-4y')(-4x+6y^2)-(6x^2-4y)(-4+12y')}\over {(-4x+6y^2)^2}}$$
Now, replacing $x=1$ $$y''(1)={{(12-4y')(-4+6y^2)-(6-4y)(-4+12y')}\over{(-4+6y^2)^2}}$$
Now, if $y=f(x)\implies y(1)=f(1,1)=0$ and $y'(1)=-1$. Replacing: $$y''(1)={{(12+4)(-4+6y^2)-(6)(-4-12)}\over{(-4+6y^2)^2}}$$
But, my question is: what can I do with $y^2$?, if $y(1)=f(1,1)=0$ can I take $y^2=0$ and $y''(1)=2$
Hints
$1.$ As the implicit equation is $f(x,y)=0$ hence
$$\begin{array}{ll} f(1,y)=2-4y+2y^3 \\ f(1,y)=2(y-1)(y^2+y-1)\\ f(1,y)=0 \\ y=1,\frac{-1 \pm \sqrt5}{2} \end{array}$$
$2.$ So just put $y=1,\frac{-1 \pm \sqrt5}{2}$ in your last equation.