Find all critical points of the following function. \( f(x, y)=\frac{x^{3}}{3}-\frac{y^{3}}{3}+9 x y \) What are the critical points? Select the correct choice below and fill in any answer boxes within your choice. A. The critical point(s) is/are (Type an ordered pair. Use a comma to separate answers as needed.) B. There are no critical points.

To find the critical points of the function \( f(x, y)=\frac{x^{3}}{3}-\frac{y^{3}}{3}+9 x y \), we first need to compute the partial derivatives and set them equal to zero. 1. Compute the partial derivatives: - \( f_x = x^2 + 9y \) (partial derivative with respect to \( x \)) - \( f_y = -y^2 + 9x \) (partial derivative with respect to \( y \)) 2. Set the partial derivatives equal to zero: - \( x^2 + 9y = 0 \) (1) - \( -y^2 + 9x = 0 \) (2) From equation (1), we can express \( y \) in terms of \( x \): \[ y = -\frac{x^2}{9} \] Substituting this into equation (2): \[ -\left(-\frac{x^2}{9}\right)^2 + 9x = 0 \] This simplifies to: \[ -\frac{x^4}{81} + 9x = 0 \] Multiplying through by 81 to eliminate the fraction gives: \[ -x^4 + 729x = 0 \] Factoring out \( x \), we get: \[ x(-x^3 + 729) = 0 \] This leads to two cases: 1. \( x = 0 \) 2. \( -x^3 + 729 = 0 \Rightarrow x^3 = 729 \Rightarrow x = 9 \) For \( x = 0 \): Substituting back into \( y = -\frac{x^2}{9} \): \[ y = 0 \quad \Rightarrow \quad (0, 0) \] For \( x = 9 \): Substituting back into \( y = -\frac{x^2}{9} \): \[ y = -\frac{9^2}{9} = -9 \quad \Rightarrow \quad (9, -9) \] Thus, the critical points are \( (0, 0) \) and \( (9, -9) \). A. The critical point(s) is/are (0, 0), (9, -9)