1. **Problem Statement:** Find all local maximum and minimum points of the function $$f(x,y) = -x^2 - 4y^2 - 2x + 8y - 1.$$\n\n2. **Formula and Rules:** To find local maxima and minima, we use the first and second derivative tests.\n- Find critical points by solving $$f_x = 0$$ and $$f_y = 0$$ where $$f_x$$ and $$f_y$$ are partial derivatives.\n- Use the second derivative test with $$D = f_{xx} f_{yy} - (f_{xy})^2$$ to classify critical points:\n - If $$D > 0$$ and $$f_{xx} > 0$$, local minimum.\n - If $$D > 0$$ and $$f_{xx} < 0$$, local maximum.\n - If $$D < 0$$, saddle point.\n\n3. **Calculate partial derivatives:**\n$$f_x = \frac{\partial}{\partial x}(-x^2 - 4y^2 - 2x + 8y - 1) = -2x - 2$$\n$$f_y = \frac{\partial}{\partial y}(-x^2 - 4y^2 - 2x + 8y - 1) = -8y + 8$$\n\n4. **Find critical points by setting derivatives to zero:**\n$$-2x - 2 = 0 \implies -2x = 2 \implies x = -1$$\n$$-8y + 8 = 0 \implies -8y = -8 \implies y = 1$$\nCritical point is at $$(-1, 1)$$.\n\n5. **Second derivatives:**\n$$f_{xx} = \frac{\partial}{\partial x}(-2x - 2) = -2$$\n$$f_{yy} = \frac{\partial}{\partial y}(-8y + 8) = -8$$\n$$f_{xy} = \frac{\partial}{\partial y}(-2x - 2) = 0$$\n\n6. **Calculate $$D$$:**\n$$D = f_{xx} f_{yy} - (f_{xy})^2 = (-2)(-8) - 0^2 = 16 > 0$$\nSince $$D > 0$$ and $$f_{xx} = -2 < 0$$, the critical point $$(-1,1)$$ is a local maximum.\n\n7. **Find the value of $$f$$ at the critical point:**\n$$f(-1,1) = -(-1)^2 - 4(1)^2 - 2(-1) + 8(1) - 1 = -1 - 4 + 2 + 8 - 1 = 4.$$\n\n**Final answer:** The function has a local maximum at $$(-1,1)$$ with value $$4$$. There are no local minima.