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$$-{2x \over y^5}$$

If 
$x^n + y^n =1
$
then
$nx^{n-1} +ny^{n-1}y' = 0$
or
$y'
=-\frac{x^{n-1}}{y^{n-1}}
$.
Differentiating again,
$\begin{array}\\
0
&=(n-1)x^{n-2}+y^{n-1}y''+(n-1)y^{n-2}(y')^2\\
&=(n-1)x^{n-2}+y^{n-1}y''+(n-1)y(\frac{x^{n-1}}{y^{n-1}})^2\\
&=y^{n-1}y''+(n-1)x^{n-2}(1+\frac{x^n}{y^{2n-3}})\\
&=y^{n-1}y''+(n-1)x^{n-2}\frac{x^n+y^{2n-3}}{y^{2n-3}}\\
\end{array}
$
so
$y''
=-\frac{(n-1)x^{n-2}\frac{x^n+y^{2n-3}}{y^{2n-3}}}{y^{n-1}}
=-\frac{(n-1)x^{n-2}(x^n+y^{2n-3})}{y^{3n-4}}
$
If $n=3$,
$y'
=-\frac{x^2}{y^2}
$
and
$y''
=-\frac{2x(x^3+y^3)}{y^5}
=-\frac{2x}{y^5}
$
since
$x^3+y^3 = 1$.
