To solve the given equation $y = \sin^{-1}(x)$ and find the second derivative $\frac{d^2y}{dx^2}$ in terms of $y$ and $x$:

1. Start with $y = \sin^{-1}(x)$.
2. Differentiate both sides with respect to $x$: $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$
3. Now, differentiate $\frac{dy}{dx}$ with respect to $x$ to find the second derivative $\frac{d^2y}{dx^2}$: $\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{1}{\sqrt{1-x^2}} \right)$
4. Use the chain rule to differentiate $\frac{1}{\sqrt{1-x^2}}$: $\frac{d}{dx} \left( \frac{1}{\sqrt{1-x^2}} \right) = \frac{d}{dx} \left(1-x^2 \right)^{-\frac{1}{2}}$
5. Apply the chain rule: $\frac{d}{dx} \left(1-x^2 \right)^{-\frac{1}{2}} = -\frac{1}{2} (1-x^2)^{-\frac{3}{2}} \cdot (-2x) = \frac{x}{(1-x^2)^{\frac{3}{2}}}$
6. Now, multiply $\frac{d^2y}{dx^2}$ by $(1-x^2)$: $(1-x^2) \frac{d^2y}{dx^2} = \frac{x}{(1-x^2)^{\frac{3}{2}}} \cdot (1-x^2)$
7. Simplify the expression: $(1-x^2) \frac{d^2y}{dx^2} = \frac{x}{(1-x^2)^{\frac{1}{2}}}$
8. The final equation is: $(1-x^2) \frac{d^2y}{dx^2} = \frac{x}{\sqrt{1-x^2}}$

Would you like more details on any part of the process or have any questions?

Here are some questions you might ask next:

1. How do you find the first derivative of an inverse function?
2. Can you explain the chain rule in differentiation?
3. How do you handle differentiating more complex functions?
4. What are some common mistakes when differentiating inverse trigonometric functions?
5. How can you apply these differentiation techniques to real-world problems?

Tip: When differentiating inverse trigonometric functions, it's crucial to remember the standard derivatives and apply the chain rule carefully to avoid mistakes.