Step-by-step Solution:

1. Understand the Implicit Equation and the Given Point The function $y = f(x)$ is defined implicitly by the equation $y^3 - 3y + x = 0$. We are given a point on this function: when $x = -10\sqrt{2}$, the corresponding $y$ value is $f(-10\sqrt{2}) = 2\sqrt{2}$.

   First, let's verify that this point satisfies the equation: $(2\sqrt{2})^3 - 3(2\sqrt{2}) + (-10\sqrt{2}) = (8 \times 2\sqrt{2}) - 6\sqrt{2} - 10\sqrt{2}$ $= 16\sqrt{2} - 6\sqrt{2} - 10\sqrt{2} = 16\sqrt{2} - 16\sqrt{2} = 0$ The point $(-10\sqrt{2}, 2\sqrt{2})$ indeed lies on the curve.

2. Find the First Derivative ($f'(x)$) We differentiate the given equation implicitly with respect to $x$: $\frac{d}{dx}(y^3 - 3y + x) = \frac{d}{dx}(0)$ $3y^2 \frac{dy}{dx} - 3 \frac{dy}{dx} + 1 = 0$ Factor out $\frac{dy}{dx}$: $(3y^2 - 3) \frac{dy}{dx} = -1$ Solve for $\frac{dy}{dx}$: $f'(x) = \frac{dy}{dx} = \frac{-1}{3y^2 - 3}$

3. Find the Second Derivative ($f''(x)$) Now, we differentiate the expression for $f'(x)$ with respect to $x$ using the chain rule. It's easier to write $f'(x) = -(3y^2 - 3)^{-1}$. $f''(x) = \frac{d^2y}{dx^2} = \frac{d}{dx} \left[ -(3y^2 - 3)^{-1} \right]$ $f''(x) = -(-1)(3y^2 - 3)^{-2} \cdot \frac{d}{dx}(3y^2 - 3)$ $f''(x) = (3y^2 - 3)^{-2} \cdot (6y \frac{dy}{dx})$ $f''(x) = \frac{6y}{(3y^2 - 3)^2} \frac{dy}{dx}$

   Now, substitute the expression for $\frac{dy}{dx}$ we found in Step 2: $f''(x) = \frac{6y}{(3y^2 - 3)^2} \left( \frac{-1}{3y^2 - 3} \right)$ $f''(x) = \frac{-6y}{(3y^2 - 3)^3}$

4. Evaluate the Second Derivative at the Given Point We need to find $f''(-10\sqrt{2})$. We use the corresponding value of $y = 2\sqrt{2}$. Substitute $y = 2\sqrt{2}$ into the expression for $f''(x)$: $f''(-10\sqrt{2}) = \frac{-6(2\sqrt{2})}{(3(2\sqrt{2})^2 - 3)^3}$

   First, calculate the term in the denominator's parenthesis: $3(2\sqrt{2})^2 - 3 = 3(4 \times 2) - 3 = 3(8) - 3 = 24 - 3 = 21$

   Now substitute this back into the expression for $f''(-10\sqrt{2})$: $f''(-10\sqrt{2}) = \frac{-12\sqrt{2}}{(21)^3}$

5. Simplify the Result We can simplify the fraction. Note that $21 = 3 \times 7$ and $12 = 3 \times 4$. $f''(-10\sqrt{2}) = \frac{-12\sqrt{2}}{(3 \times 7)^3} = \frac{-12\sqrt{2}}{3^3 \times 7^3}$ $f''(-10\sqrt{2}) = \frac{-(3 \times 4)\sqrt{2}}{3^3 \times 7^3} = \frac{-4\sqrt{2}}{3^2 \times 7^3}$

   This matches option B.

Conclusion

The second derivative at the given point is $f''(-10\sqrt{2}) = -\frac{4\sqrt{2}}{7^3 \cdot 3^2}$. Comparing this with the options provided, the correct option is B.