`"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: Substitute \( x = e \)

We start by substituting \( x = e \) into the equation.

\[ \log_e(\log_e e - \log_e y) = e^{e^2 y}(1 - \log_e e) \]

### Step 2: Simplify the equation

Since \( \log_e e = 1 \), we can simplify the equation:

\[ \log_e(1 - \log_e y) = e^{e^2 y}(1 - 1) \]

This simplifies to:

\[ \log_e(1 - \log_e y) = 0 \]

### Step 3: Solve for \( y \)

From \( \log_e(1 - \log_e y) = 0 \), we can exponentiate both sides:

\[ 1 - \log_e y = e^0 = 1 \]

This leads to:

\[ \log_e y = 0 \]

Thus, we find:

\[ y = e^0 = 1 \]

### Step 4: Differentiate the original equation

Next, we differentiate the original equation with respect to \( x \):

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

### Step 5: Substitute \( x = e \) and \( y = 1 \)

Now we substitute \( x = e \) and \( y = 1 \):

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

The right side becomes zero:

\[ \frac{1}{e} - \frac{dy}{dx} = 0 \]

### Step 6: Solve for \( \frac{dy}{dx} \)

From the equation \( \frac{1}{e} - \frac{dy}{dx} = 0 \), we can solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{1}{e} \]

### Step 7: Find \( y'(e) \)

Thus, the value of \( y'(e) \) is:

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

### Final Answer

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

---