Find the local maxima, minima, and saddle points of the function f(x,y) = x3 + 3xy2 - 15x + y3 - 15y.

Step 5: Evaluate $D(x,y)$ and $f_{xx}(x,y)$ at each critical point (continued).

For $(-2, -1)$:

$$D(-2, -1) = 36((-2)^2 + (-2)(-1) - (-1)^2)$$ $$D(-2, -1) = 36(4 + 2 - 1)$$ $$D(-2, -1) = 36(5) = 180$$

Since $D(-2, -1) > 0$, we check $f_{xx}(-2, -1)$:

$$f_{xx}(-2, -1) = 6(-2) = -12$$

Since $f_{xx}(-2, -1) < 0$ and $D(-2, -1) > 0$, $(-2, -1)$ is a local maximum.
The function value is $f(-2, -1) = (-2)^3 + 3(-2)(-1)^2 - 15(-2) + (-1)^3 - 15(-1)$

$$f(-2, -1) = -8 - 6 + 30 - 1 + 15$$ $$f(-2, -1) = -15 + 45 = 30$$

Summary of results:

• Local minima: There is a local minimum at $\boxed{(2,1)}$ with a value of $f(2,1) = \boxed{-30}$.
• Local maxima: There is a local maximum at $\boxed{(-2,-1)}$ with a value of $f(-2,-1) = \boxed{30}$.
• Saddle points: There are saddle points at $\boxed{(0, \sqrt{5})}$ with a value of $f(0, \sqrt{5}) = \boxed{-10\sqrt{5}}$ and at $\boxed{(0, -\sqrt{5})}$ with a value of $f(0, -\sqrt{5}) = \boxed{10\sqrt{5}}$.