Find the critical points of the function f(x,y) = x3 + 3xy2 - 15x + y3 - 15y.

Step 1: Find the first-order partial derivatives $f_x$ and $f_y$.
The given function is:

$$f(x,y) = x^3 + 3xy^2 - 15x + y^3 - 15y$$

Differentiate $f(x,y)$ with respect to $x$ (treating $y$ as a constant):

$$f_x = \frac{\partial}{\partial x}(x^3 + 3xy^2 - 15x + y^3 - 15y)$$ $$f_x = 3x^2 + 3y^2 - 15$$

Differentiate $f(x,y)$ with respect to $y$ (treating $x$ as a constant):

$$f_y = \frac{\partial}{\partial y}(x^3 + 3xy^2 - 15x + y^3 - 15y)$$ $$f_y = 6xy + 3y^2 - 15$$

Step 2: Find the critical points by setting $f_x = 0$ and $f_y = 0$.
Set $f_x = 0$:

$$3x^2 + 3y^2 - 15 = 0$$

Divide by 3:

$$x^2 + y^2 - 5 = 0 \quad (1)$$

Set $f_y = 0$:

$$6xy + 3y^2 - 15 = 0$$

Divide by 3:

$$2xy + y^2 - 5 = 0 \quad (2)$$

From equation (1), we have $x^2 + y^2 = 5$.
From equation (2), we have $2xy + y^2 = 5$.
Equating the two expressions for 5:

$$x^2 + y^2 = 2xy + y^2$$ $$x^2 = 2xy$$ $$x^2 - 2xy = 0$$

Factor out $x$:

$$x(x - 2y) = 0$$

This gives two cases:
Case 1: $x = 0$
Substitute $x = 0$ into equation (1):

$$0^2 + y^2 - 5 = 0$$ $$y^2 = 5 \implies y = \pm\sqrt{5}$$

This gives two critical points: $(0, \sqrt{5})$ and $(0, -\sqrt{5})$.

Case 2: $x - 2y = 0 \implies x = 2y$
Substitute $x = 2y$ into equation (1):

$$(2y)^2 + y^2 - 5 = 0$$ $$4y^2 + y^2 - 5 = 0$$ $$5y^2 - 5 = 0$$ $$5y^2 = 5 \implies y^2 = 1 \implies y = \pm 1$$

If $y = 1$, then $x = 2(1) = 2$. Critical point: $(2, 1)$.
If $y = -1$, then $x = 2(-1) = -2$. Critical point: $(-2, -1)$.
The critical points are $(0, \sqrt{5})$, $(0, -\sqrt{5})$, $(2, 1)$, and $(-2, -1)$.

Step 3: Calculate the second-order partial derivatives.

$$f_{xx} = \frac{\partial}{\partial x}(3x^2 + 3y^2 - 15) = 6x$$ $$f_{yy} = \frac{\partial}{\partial y}(6xy + 3y^2 - 15) = 6x + 6y$$ $$f_{xy} = \frac{\partial}{\partial y}(3x^2 + 3y^2 - 15) = 6y$$

Step 4: Calculate the discriminant $D(x,y)$.

$$D(x,y) = f_{xx}f_{yy} - (f_{xy})^2$$ $$D(x,y) = (6x)(6x + 6y) - (6y)^2$$ $$D(x,y) = 36x^2 + 36xy - 36y^2$$ $$D(x,y) = 36(x^2 + xy - y^2)$$

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

For $(0, \sqrt{5})$:

$$D(0, \sqrt{5}) = 36(0^2 + 0(\sqrt{5}) - (\sqrt{5})^2) = 36(0 + 0 - 5) = -180$$

Since $D(0, \sqrt{5}) < 0$, $(0, \sqrt{5})$ is a saddle point.
The function value is $f(0, \sqrt{5}) = 0^3 + 3(0)(\sqrt{5})^2 - 15(0) + (\sqrt{5})^3 - 15(\sqrt{5}) = 5\sqrt{5} - 15\sqrt{5} = -10\sqrt{5}$.

For $(0, -\sqrt{5})$:

$$D(0, -\sqrt{5}) = 36(0^2 + 0(-\sqrt{5}) - (-\sqrt{5})^2) = 36(0 + 0 - 5) = -180$$

Since $D(0, -\sqrt{5}) < 0$, $(0, -\sqrt{5})$ is a saddle point.
The function value is $f(0, -\sqrt{5}) = 0^3 + 3(0)(-\sqrt{5})^2 - 15(0) + (-\sqrt{5})^3 - 15(-\sqrt{5}) = -5\sqrt{5} + 15\sqrt{5} = 10\sqrt{5}$.

For $(2, 1)$:

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

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 minimum.
The function value is $f(2, 1) = 2^3 + 3(2)(1)^2 - 15(2) + 1^3 - 15(1) = 8 + 6 - 30 + 1 - 15 = 15 - 45 = -30$.

For $(-2, -1)$:

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