1. **Problem:** Calculate the electric field strength $\mathbf{E}$ at point $(2,3,1)$ given the potential function $$V = 8x^3 y z^2 + 16 x^2 y.$$ The electric field strength is given by $$\mathbf{E} = -\nabla V = -\text{grad } V.$$\n\n2. **Formula and rules:** The gradient operator in Cartesian coordinates is $$\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right).$$ The electric field components are $$E_x = -\frac{\partial V}{\partial x}, \quad E_y = -\frac{\partial V}{\partial y}, \quad E_z = -\frac{\partial V}{\partial z}.$$\n\n3. **Calculate partial derivatives:**\n$$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x} \left(8x^3 y z^2 + 16 x^2 y\right) = 24 x^2 y z^2 + 32 x y.$$\n$$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y} \left(8x^3 y z^2 + 16 x^2 y\right) = 8 x^3 z^2 + 16 x^2.$$\n$$\frac{\partial V}{\partial z} = \frac{\partial}{\partial z} \left(8x^3 y z^2 + 16 x^2 y\right) = 16 x^3 y z.$$\n\n4. **Evaluate at point $(2,3,1)$:**\n$$\frac{\partial V}{\partial x} \bigg|_{(2,3,1)} = 24 \times 2^2 \times 3 \times 1^2 + 32 \times 2 \times 3 = 24 \times 4 \times 3 + 192 = 288 + 192 = 480.$$\n$$\frac{\partial V}{\partial y} \bigg|_{(2,3,1)} = 8 \times 2^3 \times 1^2 + 16 \times 2^2 = 8 \times 8 + 64 = 64 + 64 = 128.$$\n$$\frac{\partial V}{\partial z} \bigg|_{(2,3,1)} = 16 \times 2^3 \times 3 \times 1 = 16 \times 8 \times 3 = 384.$$\n\n5. **Calculate electric field components:**\n$$E_x = -480, \quad E_y = -128, \quad E_z = -384.$$\n\n6. **Final answer:**\n$$\mathbf{E} = -480 \mathbf{i} - 128 \mathbf{j} - 384 \mathbf{k}.$$\n\nNote: The provided answer in the question is $-2881.98 \mathbf{j} - 288 \mathbf{k}$ which seems inconsistent with the potential given. Our calculation follows the correct gradient method for the given $V$.