Second Derivative of Implicit Function at Given Point

To find $f''(-10\sqrt{2})$, we start with the given implicit relation defining $y = f(x)$:
$y^3 - 3y + x = 0$

Differentiating both sides with respect to $x$:
$3y^2 \frac{dy}{dx} - 3\frac{dy}{dx} + 1 = 0$
$(3y^2 - 3)\frac{dy}{dx} = -1$
$\frac{dy}{dx} = -\frac{1}{3(y^2 - 1)}$

Now, differentiating implicitly once more with respect to $x$ to find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{1}{3}(y^2 - 1)^{-1}\right)$
$\frac{d^2y}{dx^2} = \frac{1}{3}(y^2 - 1)^{-2} \cdot \left(2y \frac{dy}{dx}\right) = \frac{2y}{3(y^2 - 1)^2} \frac{dy}{dx}$

Substituting $\frac{dy}{dx} = -\frac{1}{3(y^2 - 1)}$ into the expression:
$\frac{d^2y}{dx^2} = \frac{2y}{3(y^2 - 1)^2} \left(-\frac{1}{3(y^2 - 1)}\right) = -\frac{2y}{9(y^2 - 1)^3}$

We are given that at $x = -10\sqrt{2}$, $y = f(-10\sqrt{2}) = 2\sqrt{2}$.
Substituting $y = 2\sqrt{2}$ into the expression for the second derivative:
$y^2 = (2\sqrt{2})^2 = 8$
$y^2 - 1 = 8 - 1 = 7$

Hence:
$f''(-10\sqrt{2}) = -\frac{2(2\sqrt{2})}{9(7)^3} = -\frac{4\sqrt{2}}{7^3 \cdot 3^2}$

Therefore, the correct option is B.