(i) `x^(y)cdoty^(x)=1` (ii) `y=e^(x^(x))`

Answer

To solve the given problems step by step, we will analyze each part separately.

### Part (i): \( x^y \cdot y^x = 1 \)

1. **Take the logarithm of both sides**:
\[ \log(x^y \cdot y^x) = \log(1) \]
Since \(\log(1) = 0\), we have:
\[ \log(x^y) + \log(y^x) = 0 \]

2. **Use logarithmic identities**:
\[ y \log x + x \log y = 0 \]

3. **Differentiate both sides with respect to \(x\)**:
Using implicit differentiation:
\[ \frac{d}{dx}(y \log x) + \frac{d}{dx}(x \log y) = 0 \]

- For the first term, apply the product rule:
\[ \frac{dy}{dx} \log x + y \cdot \frac{1}{x} \]
- For the second term, again apply the product rule:
\[ \log y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \]

Combining these, we get:
\[ \frac{dy}{dx} \log x + \frac{y}{x} + \log y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = 0 \]

4. **Rearranging the equation**:
\[ \left(\log x + \frac{x}{y}\right) \frac{dy}{dx} = -\left(\frac{y}{x} + \log y\right) \]

5. **Solve for \(\frac{dy}{dx}\)**:
\[ \frac{dy}{dx} = -\frac{\frac{y}{x} + \log y}{\log x + \frac{x}{y}} \]

### Part (ii): \( y = e^{x^x} \)

1. **Take the logarithm of both sides**:
\[ \log y = x^x \]

2. **Differentiate both sides with respect to \(x\)**:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^x) \]

To differentiate \(x^x\), we can use the identity \(x^x = e^{x \log x}\):
\[ \frac{d}{dx}(x^x) = e^{x \log x} \left(\log x + 1\right) = x^x (\log x + 1) \]

3. **Substituting back**:
\[ \frac{1}{y} \frac{dy}{dx} = x^x (\log x + 1) \]

4. **Solve for \(\frac{dy}{dx}\)**:
\[ \frac{dy}{dx} = y \cdot x^x (\log x + 1) \]

5. **Substituting \(y\) back**:
Since \(y = e^{x^x}\), we have:
\[ \frac{dy}{dx} = e^{x^x} \cdot x^x (\log x + 1) \]

### Final Answers:
1. For part (i):
\[ \frac{dy}{dx} = -\frac{\frac{y}{x} + \log y}{\log x + \frac{x}{y}} \]

2. For part (ii):
\[ \frac{dy}{dx} = e^{x^x} \cdot x^x (\log x + 1) \]