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Решение.

  [tex]\bf f(x,y)=1+6x-xy-y^2[/tex]  

Найдём частные производные 1 порядка .

[tex]\bf \dfrac{\partial f}{\partial x}=6-y\ \ \ ,\ \ \ \dfrac{\partial f}{\partial y}=-x-2y\\\\\\\left\{\begin{array}{l}\bf 6-y=0\\\bf -x-2y=0\end{array}\right\ \ \left\{\begin{array}{l}\bf y=6\\\bf x=-2y\end{array}\right\ \ \left\{\begin{array}{l}\bf y=6\\\bf x=-12\end{array}\right\ \ \ \Rightarrow \ \ \ tochka\ (-12\, ;\ 6\, )[/tex]