91Ó°ÊÓ

First, we find the partial derivatives of the given function with respect to x, y, and z: $$\frac{\partial f}{\partial x} = yz^2 e^{xz}$$ $$\frac{\partial f}{\partial y} = ze^{xz}$$ $$\frac{\partial f}{\partial z} = yze^{xz} + 0$$ Step 2: Evaluate gradient at given points

Now, let's find the gradient of the surface at the given points by plugging in the coordinates of \(P_1(0, 2, 4)\) and \(P_2(0, -8, -1)\): At \(P_1(0, 2, 4)\): $$\nabla f_1 = \begin{bmatrix} 2(4)^2 e^{(0)(4)} \\ 4e^{(0)(2)} \\ 2(4)e^{(0)(4)} \end{bmatrix} = \begin{bmatrix} 32 \\ 4 \\ 8 \end{bmatrix}$$ At \(P_2(0, -8, -1)\): $$\nabla f_2 = \begin{bmatrix} (-8)(-1)^2 e^{0(-1)} \\ (-1)e^{0(-8)} \\ (-8)(-1)e^{0(-1)} \end{bmatrix} = \begin{bmatrix} 8 \\ -1 \\ 8 \end{bmatrix}$$ Step 3: Find the equations of tangent planes

Next, we use the point-normal form of a plane equation with the gradients as normal vectors: $$\nabla f \cdot (x - x_0, y - y_0, z - z_0) = 0$$ For \(P_1(0, 2, 4)\) and \(\nabla f_1 = \begin{bmatrix} 32 \\ 4 \\ 8 \end{bmatrix}\): $$32(x - 0) + 4(y - 2) + 8(z - 4) = 0$$ $$32x + 4y - 8 + 8z - 32 = 0$$ Then, the equation of the tangent plane at \((0, 2, 4)\) is: $$32x + 4y + 8z - 40 = 0$$ For \(P_2(0, -8, -1)\) and \(\nabla f_2 = \begin{bmatrix} 8 \\ -1 \\ 8 \end{bmatrix}\): $$8(x - 0) - 1(y + 8) + 8(z + 1) = 0$$ $$8x - y - 8 + 8z + 8 = 0$$ Then, the equation of the tangent plane at \((0, -8, -1)\) is: $$8x - y + 8z = 0$$ Thus, the equations of the tangent planes to the surface \(yz e^{xz} - 8 = 0\) at points \((0,2,4)\) and \((0,-8,-1)\) are \(32x + 4y + 8z - 40 = 0\) and \(8x - y + 8z = 0\), respectively.