91Ó°ÊÓ

Find the partial derivatives of the function with respect to \(x\) and \(y\). The partial derivative of \(f(x, y)\) with respect to \(x\) is written as \(f_x\), and similarly, the partial derivative of \(f(x, y)\) with respect to y is written as \(f_y\). We'll differentiate \(f(x, y)\) with respect to each variable. \(f_x(x, y) = \frac{\partial}{\partial x} \left( \frac{x^3}{3} - \frac{y^3}{3} + 3xy \right) = x^2 + 3y\) \(f_y(x, y) = \frac{\partial}{\partial y} \left( \frac{x^3}{3} - \frac{y^3}{3} + 3xy \right) = -y^2 + 3x\)

Now, we'll set both partial derivatives to zero and find the points where both partial derivatives are zero. \(f_x(x, y) = x^2 + 3y = 0\) \(f_y(x, y) = -y^2 + 3x = 0\)

We now have a system of two non-linear equations. To find the critical points, we'll solve the system of equations. From \(f_y(x, y) = -y^2 + 3x = 0\), we get \(y^2 = 3x\). Now, substitute \(3x = y^2\) into the first equation, \(f_x(x, y) = x^2 + 3y = 0\): \(x^2 + 3(\sqrt{3x}) = 0\) To solve this equation, we need to consider two cases: 1. \(x = 0\) In this case, \(y^2 = 3x = 0 \Rightarrow y = 0\) So, one critical point is \((0, 0)\). 2. \(x \neq 0\) In this case, we can use the substitution \(y^2 = 3x\): \(x^2 + 3(3x) = 0\) \(x(x + 9) = 0\) As we already found the solution for \(x = 0\), we now consider: \(x + 9 = 0 \Rightarrow x = -9\) Now, substituting this back into \(y^2 = 3x\), we get: \(y^2 = 3(-9) \Rightarrow y^2 = -27\) Since \(y^2\) cannot be negative, no critical points are found for this case.

The only critical point for the function \(f(x, y) = \frac{x^3}{3} - \frac{y^3}{3} + 3xy\) is \((0, 0)\).