  \textbf{4.} Spo�t�te $y'$, $y''$, $y'''$, je-li $x^2 + xy + y^2 = 3.$

      \textbf{�e�en�}
      \begin{align*}
            F(x, y) &\equiv x^2 + xy + y^2 - 3,\quad F\in{\cal C}^{\infty}\\
            \frac{\partial F}{\partial y} &= x + 2y \neq 0 \Leftrightarrow x \neq -2y\\
            &\Downarrow \text{IFT}\\
            x^2 + xy + y^2 &= 3\quad\text{na n�jak�m intervalu $J$}\\
            2x + xy' + y + 2yy' &= 0\quad\text{na $J$}\\
            y' &= -\frac{2x+y}{x+2y}\quad\text{na $J$}\\
            y'' &= -\frac{(2 + y')(x+2y) - (1 + 2y')(2x+y)}{(x+2y)^2} =\\
                &= -\frac{2x + xy' + 4y + 2yy' - 2x -4xy' -y - 2yy'}{(x+2y)^2} =\\
                &= -3\cdot\frac{y - xy'}{(x+2y)^2} = -3\cdot\frac{y + x(2x+y)/(x+2y)}{(x+2y)^2}\\
                &= -3\cdot\frac{xy + 2y^2 + 2x^2 + xy}{(x+2y)^3} = -3\cdot\frac{6}{(x+2y)^3} =\\
                &= -\frac{18}{(x+2y)^3} \quad\text{na $J$}\\
            y''' &= -18\cdot\left((x+2y)^{-3}\right)' = 54\cdot(1+2y')(x+2y)^{-4} =\\
                 &= 54\cdot\frac{\frac{x+2y - 4x - 2y}{x+2y}}{(x+2y)^4} = -\frac{162x}{(x+2y)^5}
            \quad\text{na $J.$}
      \end{align*}
