  \textbf{3.} Spo�t�te parci�ln� derivace 1. a 2. ��du pro funkci $y$, je-li:
      $$
            x^2 + 2xy - y^2 = a^2.
      $$

      \textbf{�e�en�} Definujeme funkci $F(x,y) \equiv x^2 + 2xy - y^2 - a^2.$ Tato funkce je t��dy
      $\cal C^{\infty}.$ Parci�ln� derivace $\frac{\partial F}{\partial y} = 2x - 2y \neq 0,$ pro
      $x \neq y.$ V t�chto bodech m��eme pou�it v�tu o implicitn� funkci:
      \begin{align*}
            x^2 + 2xy - y^2 - a^2 &= 0\quad \left/ \frac{\partial}{\partial x}\right.\\
            2x + 2y + 2xy' - 2yy' &= 0\\
            y' &= -\frac{x+y}{x-y}   \quad \left/ \frac{\partial}{\partial x}\right.
      \end{align*}
      \begin{align*}
            y'' &= -\frac{(x+y)'(x-y) - (x-y)'(x+y)}{(x-y)^2} =\\
                &= -\frac{(1+y')(x-y) - (1-y')(x+y)}{(x-y)^2} =\\
                &= -\frac{x - y + y'(x-y) - x - y + y'(x+y)}{(x-y)^2} =\\
                &= -\frac{2xy' - 2y}{(x-y)^2} = -\frac{-\frac{2x(x+y)}{x-y} - \frac{2y(x-y)}{x-y}}{(x-y)^2}=\\
                &= \frac{2x^2 + 2xy + 2xy - 2y^2}{(x-y)^3} = \frac{2x^2 + 4xy - 2y^2}{(x-y)^3} =\\
      \intertext{Ze zad�n� v�me, �e $2x^2 + 4xy - 2y^2 = 2a^2$, proto}
                &= \frac{2a^2}{(x-y)^3}.
      \end{align*}
