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We are asked to find \( \frac{d^2z}{dy^2} \) at \( x = 1 \), given the following relationships:

\[
\frac{dz}{dx} = 2x - 3 \quad \text{and} \quad \frac{dy}{dx} = x^2 + 1
\]

### Step-by-Step Solution:

1. **First Derivative Chain Rule Application**:
   We are required to find \( \frac{d^2z}{dy^2} \). To do this, let's use the chain rule.

   We know:
   \[
   \frac{dz}{dy} = \frac{dz}{dx} \cdot \frac{dx}{dy}
   \]
   Here, \( \frac{dx}{dy} \) is the reciprocal of \( \frac{dy}{dx} \):
   \[
   \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{x^2 + 1}
   \]

   Now, substitute this into the equation for \( \frac{dz}{dy} \):
   \[
   \frac{dz}{dy} = (2x - 3) \cdot \frac{1}{x^2 + 1}
   \]

2. **Second Derivative Application**:
   Now, we differentiate \( \frac{dz}{dy} \) with respect to \( y \) to find \( \frac{d^2z}{dy^2} \). Applying the chain rule again:
   \[
   \frac{d^2z}{dy^2} = \frac{d}{dy} \left( \frac{dz}{dy} \right) = \frac{d}{dx} \left( \frac{dz}{dy} \right) \cdot \frac{dx}{dy}
   \]
   So we need to differentiate \( \frac{dz}{dy} = \frac{(2x - 3)}{x^2 + 1} \) with respect to \( x \).

3. **Differentiate \( \frac{dz}{dy} \)**:
   Use the quotient rule:
   \[
   \frac{d}{dx} \left( \frac{2x - 3}{x^2 + 1} \right) = \frac{(x^2 + 1) \cdot 2 - (2x - 3) \cdot 2x}{(x^2 + 1)^2}
   \]

   Simplify the numerator:
   \[
   (x^2 + 1) \cdot 2 - (2x - 3) \cdot 2x = 2x^2 + 2 - (4x^2 - 6x)
   \]
   \[
   = 2x^2 + 2 - 4x^2 + 6x = -2x^2 + 6x + 2
   \]
   So, we have:
   \[
   \frac{d}{dx} \left( \frac{2x - 3}{x^2 + 1} \right) = \frac{-2x^2 + 6x + 2}{(x^2 + 1)^2}
   \]

4. **Multiply by \( \frac{dx}{dy} \)**:
   Now multiply by \( \frac{dx}{dy} = \frac{1}{x^2 + 1} \):
   \[
   \frac{d^2z}{dy^2} = \frac{-2x^2 + 6x + 2}{(x^2 + 1)^3}
   \]

5. **Evaluate at \( x = 1 \)**:
   Substitute \( x = 1 \) into the expression:
   \[
   \frac{d^2z}{dy^2} = \frac{-2(1)^2 + 6(1) + 2}{(1^2 + 1)^3} = \frac{-2 + 6 + 2}{2^3} = \frac{6}{8} = \frac{3}{4}
   \]

### Final Answer:
The second derivative \( \frac{d^2z}{dy^2} \) at \( x = 1 \) is \( \frac{3}{4} \), which corresponds to option **(b)**.

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Would you like more details on any specific part of the solution? Here are 5 related questions for deeper understanding:

1. How is the quotient rule applied when differentiating functions?
2. Can you explain more about chain rule in implicit differentiation?
3. What are the applications of second derivatives in real-world scenarios?
4. How does finding derivatives change when switching between different variables?
5. Why does the choice of evaluation point \( x = 1 \) affect the final answer?

**Tip**: Always double-check the derivatives' setup and simplifications, as small errors in the process can lead to wrong conclusions.

If dz/dx = 2x - 3, dy/dx = x^2 + 1, find d^2z/dy^2 at x = 1.

Learn how to calculate the second derivative d^2z/dy^2 at x = 1, using the chain rule and quotient rule. This calculus problem involves finding the first and second derivatives of a function with respect to y by applying key differentiation techniques. Suitable for college-level calculus students.

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