1. **State the problem:** We are given the function $f(x,y) = x^3 + y^3 - 3xy$ and want to analyze it. 2. **Formula and rules:** This is a multivariable polynomial function. To find critical points, we use partial derivatives: $$\frac{\partial f}{\partial x} = 3x^2 - 3y$$ $$\frac{\partial f}{\partial y} = 3y^2 - 3x$$ Critical points occur where both partial derivatives equal zero. 3. **Find critical points:** Set derivatives to zero: $$3x^2 - 3y = 0 \implies y = x^2$$ $$3y^2 - 3x = 0 \implies y^2 = x$$ Substitute $y = x^2$ into $y^2 = x$: $$ (x^2)^2 = x \implies x^4 = x \implies x^4 - x = 0 \implies x(x^3 - 1) = 0 $$ So $x=0$ or $x^3=1 \implies x=1$. 4. **Find corresponding $y$ values:** - For $x=0$, $y = 0^2 = 0$. - For $x=1$, $y = 1^2 = 1$. 5. **Critical points are:** $(0,0)$ and $(1,1)$. 6. **Classify critical points:** Use second derivatives: $$f_{xx} = 6x, \quad f_{yy} = 6y, \quad f_{xy} = -3$$ Compute the Hessian determinant: $$D = f_{xx} f_{yy} - (f_{xy})^2 = (6x)(6y) - (-3)^2 = 36xy - 9$$ - At $(0,0)$: $D = 36\cdot0\cdot0 - 9 = -9 < 0$, so saddle point. - At $(1,1)$: $D = 36\cdot1\cdot1 - 9 = 27 > 0$, and $f_{xx} = 6 > 0$, so local minimum. **Final answer:** The function has a saddle point at $(0,0)$ and a local minimum at $(1,1)$.