1. **Problem:** Find the unit vector normal to the surface $x^2 y + 2 x z^2 = 8$ at the point $(1,0,2)$. 2. **Formula and rules:** The normal vector to a surface defined by $F(x,y,z) = 0$ is given by the gradient vector $\nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)$. The unit normal vector is $\frac{\nabla F}{|\nabla F|}$. 3. **Step 1:** Define $F(x,y,z) = x^2 y + 2 x z^2 - 8$. 4. **Step 2:** Compute partial derivatives: $$\frac{\partial F}{\partial x} = 2 x y + 2 z^2$$ $$\frac{\partial F}{\partial y} = x^2$$ $$\frac{\partial F}{\partial z} = 4 x z$$ 5. **Step 3:** Evaluate at $(1,0,2)$: $$\frac{\partial F}{\partial x} = 2 \cdot 1 \cdot 0 + 2 \cdot 2^2 = 0 + 8 = 8$$ $$\frac{\partial F}{\partial y} = 1^2 = 1$$ $$\frac{\partial F}{\partial z} = 4 \cdot 1 \cdot 2 = 8$$ 6. **Step 4:** Gradient vector at $(1,0,2)$ is $\nabla F = (8, 1, 8)$. 7. **Step 5:** Magnitude of $\nabla F$: $$|\nabla F| = \sqrt{8^2 + 1^2 + 8^2} = \sqrt{64 + 1 + 64} = \sqrt{129}$$ 8. **Step 6:** Unit normal vector: $$\hat{n} = \frac{1}{\sqrt{129}} (8, 1, 8)$$ **Final answer:** The unit vector normal to the surface at $(1,0,2)$ is $$\boxed{\frac{8}{\sqrt{129}} \hat{i} + \frac{1}{\sqrt{129}} \hat{j} + \frac{8}{\sqrt{129}} \hat{k}}$$