Answer:
∂z/∂x = 3x^2 + 6y
∂z/∂y = 6x - 3y^2
Now, we need to solve the system of equations:
3x^2 + 6y = 0
6x - 3y^2 = 0
From the first equation, we get:
y = -x^2/2
Substituting this into the second equation, we get:
6x - 3(-x^2/2)^2 = 0
Simplifying, we get:
6x - 3x^4/4 = 0
Multiplying by 4 and rearranging, we get:
3x^4 - 24x = 0
Factoring out 3x, we get:
3x(x^3 - 8) = 0
Therefore, the critical values of x are x = 0 and x = 2.
For x = 0, we have y = 0 (from y = -x^2/2). So, one critical point is (0, 0).
For x = 2, we have y = -2. So, the other critical point is (2, -2).
To find the critical values of the function, we need to evaluate the function at each critical point:
z(0, 0) = 0^3 + 6(0)(0) - 0^3 - 1 = -1
z(2, -2) = 2^3 + 6(2)(-2) - (-2)^3 - 1 = -13
Therefore, the critical values of the function are -1 and -13.