<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] If $Dx$ is the first derivative and $Dx^2$ is the second derivative, than your first derivative is correct. For the second derivative we have:
$$
\frac{d}{dx}y'=\frac{d}{dx}\left(\frac{-x^2}{y^2}\right)$$
that, using fraction rule and chain rule for $y$, becomes:
$$\frac{-2xy^2+2x^2y(y')}{y^4}$$
substituting $y'=-x^2/y^2$ and wi a bit of algebra:
$$\frac{-2xy^3-2x^4}{y^5}=\frac{-2x(y^3+x^3)}{y^5}$$
finally, using $x^3+y^3=1$:
$$\frac{d}{dx}y'=y''=\frac{-2x}{y^5}
$$
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