1. **State the problem:** We have the function $$f(x,y,z) = x^2 y + y^2 z + z^2 x$$. We need to find all critical points, classify them using the Hessian matrix, and compute the directional derivative at the point $(1,1,1)$ in the direction of vector $\mathbf{v} = (2,-1,2)$. 2. **Find critical points:** Critical points occur where the gradient $$\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)$$ is zero. Calculate partial derivatives: $$\frac{\partial f}{\partial x} = 2xy + z^2$$ $$\frac{\partial f}{\partial y} = x^2 + 2yz$$ $$\frac{\partial f}{\partial z} = y^2 + 2zx$$ Set each to zero: $$2xy + z^2 = 0 \quad (1)$$ $$x^2 + 2yz = 0 \quad (2)$$ $$y^2 + 2zx = 0 \quad (3)$$ 3. **Solve the system:** From (1), express $z^2 = -2xy$. Substitute into (2) and (3) and analyze possible solutions. Try $x=y=z=0$: All partial derivatives are zero, so $(0,0,0)$ is a critical point. Try $x=y=z=t \neq 0$: From (1): $2t \cdot t + t^2 = 2t^2 + t^2 = 3t^2 = 0 \Rightarrow t=0$. Try $z=0$: From (1): $2xy + 0 = 0 \Rightarrow xy=0$. From (2): $x^2 + 0 = 0 \Rightarrow x=0$. From (3): $y^2 + 0 = 0 \Rightarrow y=0$. So $(0,0,0)$ again. Try $x=0$: From (1): $0 + z^2=0 \Rightarrow z=0$. From (2): $0 + 2yz=0$. From (3): $y^2 + 0=0 \Rightarrow y=0$. Again $(0,0,0)$. Try $y=0$: From (1): $0 + z^2=0 \Rightarrow z=0$. From (2): $x^2 + 0=0 \Rightarrow x=0$. From (3): $0 + 0=0$. Again $(0,0,0)$. Try $x=-y$ and $z=0$: (1): $2x(-x) + 0 = -2x^2=0 \Rightarrow x=0$. No new solutions. Hence, the only critical point is: $$\boxed{(0,0,0)}$$ 4. **Classify the critical point using Hessian matrix:** The Hessian matrix $H$ is the matrix of second derivatives: $$H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial x \partial z} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} & \frac{\partial^2 f}{\partial y \partial z} \\ \frac{\partial^2 f}{\partial z \partial x} & \frac{\partial^2 f}{\partial z \partial y} & \frac{\partial^2 f}{\partial z^2} \end{bmatrix}$$ Calculate each: $$\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(2xy + z^2) = 2y$$ $$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial y}(2xy + z^2) = 2x$$ $$\frac{\partial^2 f}{\partial x \partial z} = \frac{\partial}{\partial z}(2xy + z^2) = 2z$$ $$\frac{\partial^2 f}{\partial y \partial x} = 2x$$ $$\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}(x^2 + 2yz) = 2z$$ $$\frac{\partial^2 f}{\partial y \partial z} = \frac{\partial}{\partial z}(x^2 + 2yz) = 2y$$ $$\frac{\partial^2 f}{\partial z \partial x} = 2z$$ $$\frac{\partial^2 f}{\partial z \partial y} = 2y$$ $$\frac{\partial^2 f}{\partial z^2} = \frac{\partial}{\partial z}(y^2 + 2zx) = 2x$$ Evaluate at $(0,0,0)$: $$H(0,0,0) = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ The Hessian is the zero matrix, which is inconclusive for classification. 5. **Check nature of critical point:** Since Hessian is zero, test values around $(0,0,0)$ or use other methods. Try $f(\epsilon,0,0) = \epsilon^2 \cdot 0 + 0 + 0 = 0$. Try $f(0,\epsilon,0) = 0 + \epsilon^2 \cdot 0 + 0 = 0$. Try $f(0,0,\epsilon) = 0 + 0 + \epsilon^2 \cdot 0 = 0$. Try $f(1,1,1) = 1^2 \cdot 1 + 1^2 \cdot 1 + 1^2 \cdot 1 = 3 > 0$. Try $f(-1,-1,-1) = (-1)^2(-1) + (-1)^2(-1) + (-1)^2(-1) = 1(-1) + 1(-1) + 1(-1) = -3 < 0$. Since $f$ takes both positive and negative values near $(0,0,0)$, it is a **saddle point**. 6. **Compute directional derivative at $(1,1,1)$ in direction $\mathbf{v} = (2,-1,2)$:** First, normalize $\mathbf{v}$: $$\|\mathbf{v}\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$ Unit vector: $$\mathbf{u} = \left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right)$$ Calculate gradient at $(1,1,1)$: $$\frac{\partial f}{\partial x} = 2xy + z^2 = 2 \cdot 1 \cdot 1 + 1^2 = 2 + 1 = 3$$ $$\frac{\partial f}{\partial y} = x^2 + 2yz = 1^2 + 2 \cdot 1 \cdot 1 = 1 + 2 = 3$$ $$\frac{\partial f}{\partial z} = y^2 + 2zx = 1^2 + 2 \cdot 1 \cdot 1 = 1 + 2 = 3$$ Gradient vector: $$\nabla f(1,1,1) = (3,3,3)$$ Directional derivative: $$D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u} = 3 \cdot \frac{2}{3} + 3 \cdot \left(-\frac{1}{3}\right) + 3 \cdot \frac{2}{3} = 2 - 1 + 2 = 3$$ **Final answers:** - Critical point: $\boxed{(0,0,0)}$ (saddle point) - Directional derivative at $(1,1,1)$ in direction $(2,-1,2)$ is $\boxed{3}$.