Calculus

Find critical points by setting partial derivatives to zero: $\frac{\partial f}{\partial x} = 3x^2 - 3y = 0 \implies y = x^2$ $\frac{\partial f}{\partial y} = 3y^2 - 3x = 0 \implies x = y^2$ Substituting $y = x^2$: $x = (x^2)^2 = x^4 \implies x(x^3 - 1) = 0$. Real solutions: $x = 0 \implies y = 0$; and $x = 1 \implies y = 1$. Second derivatives: $r = \frac{\partial^2 f}{\partial x^2} = 6x, \quad s = \frac{\partial^2 f}{\partial x \partial y} = -3, \quad t = \frac{\partial^2 f}{\partial y^2} = 6y$ 1. At $(0, 0)$: $rt - s^2 = 0 - (-3)^2 = -9 < 0$ (saddle point). 2. At $(1, 1)$: $r = 6 > 0, t = 6, s = -3$. $\Delta = rt - s^2 = (6)(6) - 9 = 27 > 0$. Since $\Delta > 0$ and $r > 0$, $(1, 1)$ is a local minimum. The value of the function at $(1, 1)$ is: $f(1, 1) = 1^3 + 1^3 - 3(1)(1) = 1 + 1 - 3 = -1$