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Let's tackle both parts of the problem.

### The given curve is:
\[
x \ln y = 2x + 1
\]

#### (i) Show that \(\frac{dy}{dx} = -\frac{y}{x^2}\).

We start with the equation \(x \ln y = 2x + 1\) and differentiate both sides with respect to \(x\), applying implicit differentiation:

1. **Differentiate the left-hand side** \(x \ln y\):
   - Use the product rule: \(\frac{d}{dx}(x \ln y) = \ln y \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(\ln y)\).
   - \(\frac{d}{dx}(x) = 1\) and \(\frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx}\).
   - Thus, \(\frac{d}{dx}(x \ln y) = \ln y + \frac{x}{y} \cdot \frac{dy}{dx}\).

2. **Differentiate the right-hand side** \(2x + 1\):
   - The derivative of \(2x + 1\) is simply 2.

Now, we equate the derivatives of both sides:
\[
\ln y + \frac{x}{y} \cdot \frac{dy}{dx} = 2
\]

3. **Solve for \(\frac{dy}{dx}\)**:
   - First, rearrange the equation: \(\frac{x}{y} \cdot \frac{dy}{dx} = 2 - \ln y\).
   - Multiply both sides by \(\frac{y}{x}\) to isolate \(\frac{dy}{dx}\):
   \[
   \frac{dy}{dx} = \frac{y(2 - \ln y)}{x}
   \]
   - Substitute the original equation \(x \ln y = 2x + 1\), or \(\ln y = \frac{2x+1}{x}\), into the expression.
   - After simplification, you get \(\frac{dy}{dx} = -\frac{y}{x^2}\).

This proves part (i).

#### (ii) Find the equation of the tangent line at the point where \(y = 1\).

The point where \(y = 1\):
- From the given equation \(x \ln y = 2x + 1\), substitute \(y = 1\), which gives \(\ln(1) = 0\), so the equation becomes \(0 = 2x + 1\), or \(x = -\frac{1}{2}\).

Now, we need to find the slope of the tangent line at this point:
1. From part (i), we know \(\frac{dy}{dx} = -\frac{y}{x^2}\).
2. At the point \((x, y) = \left(-\frac{1}{2}, 1\right)\), substitute \(x = -\frac{1}{2}\) and \(y = 1\):
   \[
   \frac{dy}{dx} = -\frac{1}{\left(-\frac{1}{2}\right)^2} = -\frac{1}{\frac{1}{4}} = -4
   \]
   So, the slope of the tangent is \(-4\).

The equation of the tangent line is:
\[
y - y_1 = m(x - x_1)
\]
where \((x_1, y_1) = \left(-\frac{1}{2}, 1\right)\) and \(m = -4\):
\[
y - 1 = -4\left(x + \frac{1}{2}\right)
\]
Simplifying:
\[
y - 1 = -4x - 2
\]
\[
y = -4x - 1
\]

Rearranging into the form \(ax + by + c = 0\):
\[
4x + y + 1 = 0
\]

Thus, the equation of the tangent line is \(4x + y + 1 = 0\), which matches the given answer.

---

Would you like further details on any step? Here are 5 related questions to deepen understanding:

1. How do you apply the product rule when differentiating implicit functions?
2. Why is it necessary to use implicit differentiation for this type of equation?
3. What are the geometric interpretations of the tangent line and its slope at a given point?
4. Can we use this method for other logarithmic functions? How?
5. How does the product rule differ from the chain rule in implicit differentiation?

**Tip**: When working with logarithmic equations, always check the domain of the function to ensure valid operations.

The equation of a curve is x ln y = 2x + 1. (i) Show that dy/dx = -y/x^2. (ii) Find the equation of the tangent to the curve at the point where y = 1, giving your answer in the form ax + by + c = 0.

Learn how to differentiate the curve x ln y = 2x + 1 implicitly and find the equation of the tangent line at the point where y = 1. This problem covers key calculus concepts such as implicit differentiation, logarithmic functions, and the product rule. Suitable for advanced high school calculus students.

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