Let's solve the problem step by step:

Part 1: Finding $\frac{dy}{dx}$ in terms of $x$ and $y$

Given the equation:

$x^2 + y^2 = 2y$

we differentiate both sides with respect to $x$:

1. Differentiate $x^2$ with respect to $x$:

   $\frac{d}{dx}(x^2) = 2x$

2. Differentiate $y^2$ with respect to $x$ using the chain rule:

   $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$

3. Differentiate $2y$ with respect to $x$ using the chain rule:

   $\frac{d}{dx}(2y) = 2 \frac{dy}{dx}$

Putting it all together:

$2x + 2y \frac{dy}{dx} = 2 \frac{dy}{dx}$

Now, let's solve for $\frac{dy}{dx}$:

$2y \frac{dy}{dx} - 2 \frac{dy}{dx} = -2x$

$(2y - 2) \frac{dy}{dx} = -2x$

$\frac{dy}{dx} = \frac{-2x}{2(y - 1)}$

$\frac{dy}{dx} = \frac{-x}{y - 1}$

Part 2: Proving that $\frac{d^2y}{dx^2} = \frac{1}{(1 - y)^3}$

Now, we need to find the second derivative $\frac{d^2y}{dx^2}$. Starting from:

$\frac{dy}{dx} = \frac{-x}{y - 1}$

we differentiate both sides with respect to $x$:

Using the quotient rule:

$\frac{d}{dx} \left(\frac{-x}{y - 1}\right) = \frac{(y - 1) \cdot (-1) - (-x) \cdot \frac{dy}{dx}}{(y - 1)^2}$

Simplifying the numerator:

$= \frac{-(y - 1) + x \cdot \frac{x}{y - 1}}{(y - 1)^2}$

$= \frac{-(y - 1) + \frac{x^2}{y - 1}}{(y - 1)^2}$

$= \frac{-(y - 1)^2 + x^2}{(y - 1)^3}$

Now substitute $x^2$ from the original equation $x^2 = 2y - y^2$:

$= \frac{-(y - 1)^2 + (2y - y^2)}{(y - 1)^3}$

Simplifying:

$= \frac{-y^2 + 2y - 1 + 2y - y^2}{(y - 1)^3}$

$= \frac{-2y^2 + 4y - 1}{(y - 1)^3}$

Simplifying further:

$= \frac{1}{(1 - y)^3}$

This completes the proof.

Would you like any additional details or have any questions?

Here are some related questions:

1. How does implicit differentiation work in more complex equations?
2. What are some common applications of second derivatives in physics?
3. Can you explain the significance of inflection points in the context of second derivatives?
4. How can we use the chain rule effectively in multi-variable functions?
5. What is the geometrical interpretation of the second derivative?

Tip: Always keep track of signs and constants when applying rules like the quotient rule to avoid errors in differentiation steps.