`"If "log_(e)(log_(e) x-log_(e)y)=e^(x^(2_(y)))(1-log_(e)x)," then find the value of "y'(e).`

Explanation

To solve the equation \( \log_e(\log_e x - \log_e y) = e^{x^2 y}(1 - \log_e x) \) and find the value of \( y'(e) \), we will follow these steps:

### Step 1: Differentiate both sides
We start by differentiating both sides of the equation with respect to \( x \).

\[ \frac{d}{dx} \left( \log_e(\log_e x - \log_e y) \right) = \frac{d}{dx} \left( e^{x^2 y}(1 - \log_e x) \right) \]

### Step 2: Apply the chain rule on the left side
Using the chain rule on the left side:

\[ \frac{1}{\log_e x - \log_e y} \cdot \left( \frac{1}{x} - \frac{1}{y} \cdot y' \right) \]

### Step 3: Differentiate the right side
For the right side, we will use the product rule:

\[ \frac{d}{dx} \left( e^{x^2 y} \right) \cdot (1 - \log_e x) + e^{x^2 y} \cdot \frac{d}{dx}(1 - \log_e x) \]

Calculating the derivative of \( e^{x^2 y} \):

\[ \frac{d}{dx}(e^{x^2 y}) = e^{x^2 y} \cdot (2xy + x^2 y') \]

And the derivative of \( (1 - \log_e x) \):

\[ \frac{d}{dx}(1 - \log_e x) = -\frac{1}{x} \]

### Step 4: Combine the derivatives
Putting it all together, we have:

\[ \frac{1}{\log_e x - \log_e y} \left( \frac{1}{x} - \frac{1}{y} y' \right) = e^{x^2 y} \left( (2xy + x^2 y')(1 - \log_e x) - \frac{e^{x^2 y}}{x} \right) \]

### Step 5: Substitute \( x = e \)
Now we substitute \( x = e \):

\[ \log_e e = 1 \quad \text{and} \quad \log_e y = 0 \Rightarrow y = 1 \]

### Step 6: Evaluate the derivatives
Substituting \( x = e \) and \( y = 1 \):

\[ \frac{1}{1 - 0} \left( \frac{1}{e} - \frac{1}{1} y' \right) = e^{e^2} \left( (2e \cdot 1 + e^2 y')(1 - 1) - \frac{e^{e^2}}{e} \right) \]

This simplifies to:

\[ \frac{1}{e} - y' = 0 \]

### Step 7: Solve for \( y' \)
Thus, we find:

\[ y' = \frac{1}{e} \]

### Final Answer
The value of \( y'(e) \) is \( \frac{1}{e} \).

---