<s>[INST] Trouble finding solution via implicit differentiation I'm currently using the book University Calculus: Alternate Edition (Hass et al., 2008) to study calculus. I was studying implicit differentiation and ran into an exercise problem that I was having trouble solving. The question is:

Use implicit differentiation to find $dy/dx$ and $d^2y/dx^2$ for:
$$x^{2/3} + y^{2/3} = 1$$

I was able to find $y'$ fairly simply as:
$$y' = -\left( \dfrac{y}{x} \right)^{1/3}$$
but I'm having trouble finding $y''$.
My approach is as follows:
$$
\begin{align}
y'' & = \frac{d}{dx}y' \\
& = \frac{d}{dx} \left( \frac{-y^{1/3}}{x^{1/3}} \right) \\
& = \frac{(-y^{1/3})'x^{1/3} - (-y^{1/3})(x^{1/3})'}{(x^{1/3})^2} \\
& = \frac{-\frac{1}{3}y^{-2/3} y' x^{1/3} + y^{1/3} \frac{1}{3}x^{-2/3}}{x^{2/3}} \\
& = \dfrac{\dfrac{-y^{-2/3} y' x^{1/3}}{3}  + \dfrac{y^{1/3}}{3}}{x^{4/3}} \\
& = \dfrac{\left( -y^{-2/3} \times \dfrac{-y^{1/3}}{\phantom{-}x^{1/3}} \times x^{1/3}\right) + y^{1/3}}{3x^{4/3}} \\
& = \dfrac{y^{1/3} + y^{-1/3}}{3x^{4/3}}
\end{align}
$$
However, the textbook solution states that the correct answer for $y''$ is:
$$y'' = \dfrac{x^{2/3}y^{-1/3} - y^{1/3}}{3x^{4/3}}$$
and I'm having trouble reverse engineering where I went wrong. Any tips or pointers are appreciated. Thanks in advance.
 [/INST] Using the Quotient Rule in finding higher derivatives from implicit differentiation can create opportunities for errors to "creep into" the calculations.  Starting from $ \ x^{2/3} + y^{2/3} \ = \ 1 \ \ , $ you would have produced $ \ \frac23·x^{-1/3} + \frac23·y^{-1/3}·y' \ = \ 0 \ \ $ to  obtain $ \ y' \ = \ -\frac{y^{1/3}}{x^{1/3}} \ \ . $  We will want to hold that aside for what follows.
We can return to the equation from which we found the first derivative and differentiate it implicitly one more:
$$ \ \frac{d}{dx} \ \left[ \ \frac23·x^{-1/3} \ + \ \frac23·y^{-1/3}·y' \ \right] \ \ = \ \ \frac{d}{dx} \  [ \ 0 \ ] \ $$
$$ \Rightarrow \ \ \frac23·\left(-\frac13 \right)·x^{-4/3} \ + \ \underbrace{\left[ \  \frac23· \left(-\frac13 \right)·y^{-4/3}·y' \ \right] · y' \ + \ \frac23·y^{-1/3}·y''}_{\text{by Product Rule}}   $$
$$ = \ \ -\frac29 ·x^{-4/3} \ - \   \  \frac29· y^{-4/3}·(y')^2 \  \ + \ \frac23·y^{-1/3}·y'' \ \ = \ \ 0 $$
$$ \Rightarrow \ \ y'' \ \ = \ \ \frac{x^{-4/3} \ + \   \   y^{-4/3}·(y')^2}{3·y^{-1/3}} \ · \ \frac{x^{4/3}y^{4/3}}{x^{4/3}y^{4/3}} \ \ = \ \ \large{\frac{y^{4/3} \ + \   \   x^{ 4/3}·\left(-\frac{y^{1/3}}{x^{1/3}} \right)^2}{3·x^{4/3}·y }} $$
$$ = \ \  \frac{y^{4/3} \ + \   \   x^{ 2/3}·y^{ 2/3}} {3·x^{4/3}·y }  \ \ = \ \ \frac{y^{1/3} \ + \   \   x^{ 2/3}·y^{-1/3}} {3·x^{4/3} } \ \ . $$
So by this alternative calculation, I agree with John Omielan and believe that the textbook solution either contains a "typo" or the solver lost track of a "minus-sign".  (They could also have made use of the original curve equation -- that of an astroid -- to write
$$ \ y'' \ \   = \ \ \frac{y^{1/3} \ + \   \   x^{ 2/3}·y^{-1/3}} {3·x^{4/3} } \ · \ \frac{y^{1/3}}{y^{1/3}} \ \ = \ \ \frac{y^{2/3} \ + \   \   x^{ 2/3}} {3·x^{4/3}· y^{1/3}  } \ \ = \ \ \frac{1} {3·x^{4/3}· y^{1/3}  } \ \ . \ ) $$
</s>