$$ f ^ { \prime \prime } ( x ) = ( \frac { e ^ { x } x - e ^ { x } + 1 } { e ^ { x } x - x } ) ^ { \prime } $$

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To compute the second derivative $$ f''(x) $$ given by:

$$
f''(x) = \left( \frac{e^x x - e^x + 1}{e^x x - x} \right)',
$$

we'll follow these steps:

1. **Let’s Define the Numerator and Denominator:**
   $$
   N = e^x x - e^x + 1
   $$
   $$
   D = e^x x - x
   $$

2. **Compute the First Derivatives $$ N' $$ and $$ D' $$:**
   $$
   N' = \frac{d}{dx}(e^x x) - \frac{d}{dx}(e^x) + \frac{d}{dx}(1) = e^x x + e^x - e^x = e^x x
   $$
   $$
   D' = \frac{d}{dx}(e^x x) - \frac{d}{dx}(x) = e^x x + e^x - 1 = e^x(x + 1) - 1
   $$

3. **Apply the Quotient Rule:**
   The quotient rule states that:
   $$
   \left( \frac{N}{D} \right)' = \frac{N' D - N D'}{D^2}
   $$
   Substituting the values:
   $$
   f''(x) = \frac{e^x x (e^x x - x) - (e^x x - e^x + 1)(e^x(x + 1) - 1)}{(e^x x - x)^2}
   $$

4. **Simplify the Numerator:**
   Expanding the terms in the numerator:
   $$
   e^x x (e^x x - x) = e^{2x} x^2 - e^x x^2
   $$
   $$
   (e^x x - e^x + 1)(e^x(x + 1) - 1) = e^{2x} x^2 + e^{2x} x - e^x x - e^{2x} x - e^{2x} + e^x + e^x x + e^x - 1 = e^{2x} x^2 - e^{2x} + 2e^x - 1
   $$
   Thus, the numerator becomes:
   $$
   e^{2x} x^2 - e^x x^2 - e^{2x} x^2 + e^{2x} - 2e^x + 1 = e^{2x} - e^x (x^2 + 2) + 1
   $$

5. **Final Expression for $$ f''(x) $$:**
   $$
   f''(x) = \frac{e^{2x} - e^x (x^2 + 2) + 1}{x^2 (e^x - 1)^2}
   $$

**Final Answer:**
$$
f''(x) = \frac{\,e^{2x} - e^{x}\left(x^{2} + 2\,\right) + 1\,}{\,x^{2}\left(e^{x} - 1\right)^{2}\,}
$$
The second derivative $$ f''(x) $$ is: $$ f''(x) = \frac{e^{2x} - e^{x}(x^2 + 2) + 1}{x^2 (e^x - 1)^2} $$

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