\begin{frame}
  \frametitle{Implicit Differentiation}

  \begin{exampleblock}{}
    Find $y'$ where $x^3 + y^3 = 6xy$.
    \begin{talign}
      y' = \frac{ 2y - x^2}{y^2 - 2x}
    \end{talign}
    
    At what point in the first quadrant is the tangent horizontal?
    \pause\medskip
    
    In the first quadrant $x > 0$ and $y > 0$\pause, and
    \begin{talign}
      &2y - x^2 = 0 \mpause[1]{\quad\implies\quad y = \frac{x^2}{2}}
       \mpause[2]{\quad\implies\quad x^3 + (\frac{x^2}{2})^3 = 6x\frac{x^2}{2}} \\
      &\mpause[3]{\quad\implies\quad \frac{x^6}{8} - 2x^3 = 0} 
      \mpause[4]{\quad\implies\quad x^3(\frac{x^3}{8} - 2) = 0} 
    \end{talign}
    \pause\pause\pause\pause\pause
    Since $x > 0$, we get $x = \sqrt[3]{16}$. \pause Then the point is
    \begin{talign}
      (\sqrt[3]{16},\sqrt[3]{32})
    \end{talign}
  \end{exampleblock}
  \vspace{10cm}
\end{frame}