Paragraph:
Consider the functions defined implicitly by the equation \(y^3-3 y+x=0\) on various intervals in the real line. If \(x \in(-\infty,-2) \cup(2, \infty)\), the equation implicitly defines a unique real valued differentiable function \(y=f(x)\). If \(x \in(-2,2)\), the equation implicitly defines a unique real valued differentiable function \(y=g(x)\), satisfying \(g(0)=0\).Question:
If \(f(-10 \sqrt{2})=2 \sqrt{2}\), then \(f^{\prime \prime}(-10 \sqrt{2})\) is equal to

Given, \(y^3-3 y+x=0 \Rightarrow 3 y^2 \frac{d y}{d x}-3 \frac{d y}{d x}+1=0\)
\[
\Rightarrow \quad 3 y^2\left(\frac{d^2 y}{d x^2}\right)+6 y\left(\frac{d y}{d x}\right)^2-3 \frac{d^2 y}{d x^2}=0
\]
On substituting \(x=-10 \sqrt{2}, y=2 \sqrt{2}\) in Eq. (i), we get
\[
3(2 \sqrt{2})^2 \cdot \frac{d y}{d x}-3 \cdot \frac{d y}{d x}+1=0 \Rightarrow \frac{d y}{d x}=\frac{-1}{21}
\]
Again on substituting \(x=-10 \sqrt{2}, y=2 \sqrt{2}\) in Eq. (ii), we get
\[
\begin{aligned}
& 3(2 \sqrt{2})^2 \frac{d^2 y}{d x^2}+6(2 \sqrt{2}) \cdot\left(\frac{-1}{21}\right)^2-3 \cdot \frac{d^2 y}{d x^2}=0 \\
& \Rightarrow \quad 21 \cdot \frac{d^2 y}{d x^2}=-\frac{12 \sqrt{2}}{(21)^2} \Rightarrow \frac{d^2 y}{d x^2}=-\frac{12 \sqrt{2}}{(21)^3}=-\frac{4 \sqrt{2}}{7^3 \cdot 3^2}
\end{aligned}
\]