<s>[INST] Second derivative of $x^3+y^3=1$ using implicit differentiation I need to find the $D_x^2y$ of     $x^3+y^3=1$ using implicit differentiation
So,
$$
x^3 + y^3 =1 \\
3x^2+3y^2 \cdot D_xy = 0 \\
3y^2 \cdot D_xy= -3x^2 \\
D_xy = - {x^2 \over y^2}
$$
Now I need to find the $D_x^2y$.
I am pretty sure that means the second derivative.
How would I do it to find the second derivative? apparently, it is supposed to be$$-{2x \over y^5}$$
 [/INST] Throughout this solution, I will use the Leibniz notation $\frac{dy}{dx}$ to represent the derivative of $y$ with respect to $x.$ Notice that the second derivative is notated $\frac{d^{2}y}{dx^{2}}.$ We begin as you did - find the first derivative.
Given: $x^{3} + y^{3} = 1$
$3x^{2} + 3y^{2} \cdot \frac{dy}{dx} = 0$ (*)
$\frac{dy}{dx} = -\frac{x^{2}}{y^{2}}$ (**)
We differentiate the (*) equation with respect to $x.$ This yields the following:
$6x + \frac{dy}{dx} \cdot (6y \cdot \frac{dy}{dx}) + (3y^{2}) \cdot (\frac{d^{2}y}{dx^{2}}) = 0$
$6x + 6y \cdot (\frac{dy}{dx})^{2} + 3y^2 \cdot (\frac{d^{2}y}{dx^{2}}) = 0$
$3y^2 \cdot (\frac{d^{2}y}{dx^{2}}) = -6x - 6y \cdot (\frac{dy}{dx})^{2}$
Substituting the (**) equation for $\frac{dy}{dx}$ here, we find our answer:
$\frac{d^{2}y}{dx^{2}} = \frac{-2x - 2y \cdot (-\frac{x^{2}}{y^{2}})^{2}}{y^{2}}$
$\frac{d^{2}y}{dx^{2}} = \boxed{-\frac{2x}{y^{2}}(1 + \frac{x^{3}}{y^{3}})}.$
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