1. **Problem statement:** Find and classify the critical points of the function $$f(x,y) = x^3 - 8y^3 - 12xy + 5.$$\n\n2. **Step 1: Find the partial derivatives.**\nWe calculate the first-order partial derivatives to find critical points where both derivatives are zero.\n$$f_x = \frac{\partial}{\partial x}(x^3 - 8y^3 - 12xy + 5) = 3x^2 - 12y$$\n$$f_y = \frac{\partial}{\partial y}(x^3 - 8y^3 - 12xy + 5) = -24y^2 - 12x$$\n\n3. **Step 2: Set the partial derivatives equal to zero to find critical points.**\n$$3x^2 - 12y = 0 \implies 3x^2 = 12y \implies y = \frac{3x^2}{12} = \frac{x^2}{4}$$\n$$-24y^2 - 12x = 0 \implies -24y^2 = 12x \implies x = -2y^2$$\n\n4. **Step 3: Substitute $y = \frac{x^2}{4}$ into $x = -2y^2$.**\n$$x = -2\left(\frac{x^2}{4}\right)^2 = -2\left(\frac{x^4}{16}\right) = -\frac{x^4}{8}$$\nRearranged: $$x + \frac{x^4}{8} = 0$$\nMultiply both sides by 8 to clear denominator: $$8x + x^4 = 0$$\nFactor: $$x(8 + x^3) = 0$$\nSo, $$x = 0$$ or $$8 + x^3 = 0 \implies x^3 = -8 \implies x = -2$$\n\n5. **Step 4: Find corresponding $y$ values.**\n- For $x=0$: $$y = \frac{0^2}{4} = 0$$\n- For $x=-2$: $$y = \frac{(-2)^2}{4} = \frac{4}{4} = 1$$\n\n6. **Step 5: Critical points are $(0,0)$ and $(-2,1)$.**\n\n7. **Step 6: Classify critical points using the second derivative test.**\nCalculate second derivatives:\n$$f_{xx} = \frac{\partial}{\partial x}(3x^2 - 12y) = 6x$$\n$$f_{yy} = \frac{\partial}{\partial y}(-24y^2 - 12x) = -48y$$\n$$f_{xy} = \frac{\partial}{\partial y}(3x^2 - 12y) = -12$$\n\n8. **Step 7: Compute the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$ at each critical point.**\n- At $(0,0)$:\n$$f_{xx} = 6(0) = 0$$\n$$f_{yy} = -48(0) = 0$$\n$$f_{xy} = -12$$\n$$D = (0)(0) - (-12)^2 = -144 < 0$$\nSince $D < 0$, $(0,0)$ is a saddle point.\n\n- At $(-2,1)$:\n$$f_{xx} = 6(-2) = -12$$\n$$f_{yy} = -48(1) = -48$$\n$$f_{xy} = -12$$\n$$D = (-12)(-48) - (-12)^2 = 576 - 144 = 432 > 0$$\nSince $D > 0$ and $f_{xx} < 0$, $(-2,1)$ is a local maximum.\n\n**Final answer:**\n- Critical points: $(0,0)$ (saddle point), $(-2,1)$ (local maximum).