To find the absolute extrema of the function $f(x, y) = x^3 - 6xy - y^3$ on the closed and bounded region $R = \{(x, y): -3 \leq x \leq 3, -3 \leq y \leq 3 \}$, we follow these steps:

1. Find the critical points in the interior of $R$

We first compute the partial derivatives of $f(x, y)$ and set them equal to zero to find the critical points.

a) Compute the partial derivatives:

$f_x(x, y) = \frac{\partial}{\partial x}(x^3 - 6xy - y^3) = 3x^2 - 6y$ $f_y(x, y) = \frac{\partial}{\partial y}(x^3 - 6xy - y^3) = -6x - 3y^2$

b) Set the partial derivatives equal to zero to find the critical points:

$3x^2 - 6y = 0 \quad \text{(1)}$ $-6x - 3y^2 = 0 \quad \text{(2)}$

From equation (1), solve for $y$:

$3x^2 - 6y = 0 \implies y = \frac{x^2}{2} \quad \text{(3)}$

Substitute equation (3) into equation (2):

$-6x - 3\left(\frac{x^2}{2}\right)^2 = 0$

This simplifies to:

[ -6x - 3 \left(\frac{x^4.append(String)}} \ti}