      \textbf{2.} Je d�n vztah $x^2 + 2y^2 + 3z^2 + xy - z - 9 = 0$ a bod $[1,-2,1]$:
      \begin{itemize}
            \item Doka�te, �e t�mto vztahem je definov�na hladk� funkce $z = z(x,y)$
                  v jist�m okol� $U$ bodu $[1,-2]$, pro kterou plat� $z(1,-2) = 1$;
            \item ur�ete $\frac{\partial z}{\partial x}$, $\frac{\partial z}{\partial y}$
                  v okol� $U$; napi�te rovnici te�n� roviny ke grafu funkce $z = z(x,y)$
                  v bode $[1,-2]$.
      \end{itemize}

      \textbf{�e�en�} Ozna�me $G = x^2 + 2y^2 + 3z^2 + xy - z - 9$. Plat�, �e $G(1,-2,1) = 0$
      a $G\in\mathcal{C}^1$. Plat� tak�, �e
      $$
            \frac{\partial G}{\partial z} = 6z - 1 \neq 0 \quad\text{pro } (x,y,z) = (1,-2,1),
      $$
      tedy m��eme pou�it v�tu o implicitn� funkci a plat�:
      \begin{gather*}
            z = z(x, y) \quad\text{na }J\\
            x^2 + 2y^2 + 3z^2 + xy - z - 9 = 0\\
            z \in \mathcal C ^1.
      \end{gather*}
      To, �e $z(1,-2) = 1$ plyne z toho, �e $z$ je funkce a $G\bigl(x, y, z(x,y)\bigr) = 0$.
      Parci�ln� derivace funkce $z$ vypo�teme ze vzorce:
      \begin{align*}
            \frac{\partial z}{\partial x} &= -\frac   {\frac{\partial G}{\partial x}}
                                          {\frac{\partial G}{\partial z}} =
            -\frac{2x + y}{6z - 1}\\
            \frac{\partial z}{\partial y} &= -\frac   {\frac{\partial G}{\partial y}}
                                          {\frac{\partial G}{\partial z}} =
            -\frac{4y + x}{6z - 1}\\
            \frac{\partial z}{\partial y}(1,-2,1) &= -\frac{-8+1}{5} = \frac 75\\
            \frac{\partial z}{\partial x}(1,-2,1) &= 0
      \end{align*}
      Rovnice te�n� roviny ke grafu funkce $z$ v bod� $(x,y)=(1,-2)$: $p(x,y) = \frac{7y}{5}$.
