Finding the Derivative of a Derivative 
Let $x^3+y^3=9$. Find $y''(x)$ at the point $(2,1)$.

I keep getting 

$3x^2+(3y^2)y'=0$

as the first derivative then simplify that down to

$-3x^2/(3y^2).$

But after that I keep getting messed up. If you could leave me step by step instructions and how you got your answer I'd be forever grateful. Thanks for your time.
 A: So $y^3=9-x^3$ and therefore $y(x)=(9-x^3)^{1/3}$ so $$y'(x)=-x^2(9-x^3)^{-2/3}$$ and $$y''(x)=\frac{-2x}{(9-x^3)^{2/3}}-\frac{-2x^4}{(9-x^3)^{5/3}}$$ which gives $$y''(2)=-36$$.
A: First, find first derivative
$$
3x^2 + 3y^2y' = 0 \implies y' = -\frac {x^2}{y^2} = -4
$$
Now, find second derivative
$$
\left( x^2 + y^2 y'\right )' = 2x + 2y (y')^2 + y^2 y'' = 0 \implies y'' = -\frac {2\left( x + y(y')^2\right )}{y^2} = -2(2 + 16) = -36
$$
A: $$x^3+y^3=9\\
y^3=9-x^3\\
3y^2y'=-3x^2\\
6y(y')^2+3y^2y''=-6x\\
6y(\frac{-x^2}{y^2})^2+3y^2y''=-6x\\
\frac{6x^4}{y^3}+3y^2y''=-6x\\
y''=\frac{-2xy^3-2x^4}{y^5}$$
At (2,1) $y''=-36$
A: Your way will work nicely. But I am not good at division. We have $3x^2+3y^2 y'=0$, or more simply 
$$x^2+y^2 y'=0.$$
Now don't hesitate, differentiate (again). We get
$$2x+y^2y'' +2y(y')^2=0.$$
Now replace $y'$ by $-\frac{x^2}{y^2}$, and solve for $y''$. 
Remark: If I am interested in a particular point, such as the point $(2,1)$, I would compute $y'$ at $(2,1)$ before solving. 
A: You have $3x^2+3y^2y^{\prime}=0$, so substituting the values of x and y gives $y^{\prime}=-4$ at the point $(2,1)$.
Differentiating again gives $6x+3y^2y^{\prime\prime}+(6yy^{\prime})y^{\prime}=0$, so substituting $x=2$, $y=1$, $y^{\prime}=-4$ gives
$12+3y^{\prime\prime}+6(16)=0$, so $y^{\prime\prime}=-36$.
