Second Derivative of Implicit Function My original equation started out as: $x^2 -4xy + 4y + 8 = 0$
I found the first derivative to be $2x -4x\frac{dy}{dx} - 4y + 4\frac{dy}{dx} = 0$, which I am assuming is right.
I then need to find the second derivative in order to find if the turning points on the curve are a minimum or a maximum.  I re-arranged my previous equation to get:
$$\frac{dy}{dx} = \frac{4y - 2x}{4-4x}$$
And then I used the quotient rule and subbed in $\frac{dy}{dx}$ where it occured to get:
$$\frac{d^2y}{dx^2} = \frac{8}{(4-4x)^2}[\frac{8y - 8xy - 4x + 4x^2}{4 - 4x} - 1 + 2y]$$
I can leave it in terms of $x$ and $y$ since I can find $y$ from the original equation.  Have I done this correctly?

We are asked to find the second derivative using implicit differentiation for:
$$x^2 -4xy + 4y + 8 = 0 \tag 1$$
Differentiating $(1)$:
$$2 x -4(y + x y') + 4 y' = 0 \tag 2$$
Solving for $y'$ in $(2)$ while noting we can divide out a $2$:
$$y' = \dfrac{x-2y}{2(x-1)} \tag 3$$
Next, we will reuse $(2)$ to find the second derivative, differentiating:
$$2 - 4(y' + y' + x y'') + 4 y'' = 0$$
Simplifying while noting we can divide out a $2$:
$$2(1-x)y'' = -1 + 4 y' = -1 + 4\left(\dfrac{x-2y}{2(x-1)}\right)=-\dfrac{1+x -4y}{(x-1)}$$
Solving for $y''$:
$$y'' = -\dfrac{1+x -4y}{2(x-1)^2}$$
