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.

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
$$ \ y'' \ \   = \ \ \frac{y^{1/3} \ + \   \   x^{ 2/3}·y^{-1/3}} {3·x^{4/3} } \ · \ \frac{y^{1/3}}{y^{1/3}} \ \ = \ \ \frac{y^{2/3} \ + \   \   x^{ 2/3}} {3·x^{4/3}· y^{1/3}  } \ \ = \ \ \frac{1} {3·x^{4/3}· y^{1/3}  } \ \ . \ ) $$
