$$f(x) = \int\limits_0^x {{e^{x - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $$

$$f(x) = {e^x}\int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $$

$${e^{ - x}}f(x) = \int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1)} $$

Differentiate on both side

$${e^{ - x}}f'(x) + ( - f(x){e^{ - x}}) = {e^{ - x}}f'(x) - 2x + 1$$

$$f(x) = {e^x}(2x - 1)$$

$$f'(x) = {e^x}(2) + {e^x}(2x - 1)$$

$$ = {e^x}(2x + 1)$$

$$x = - {1 \over 2}$$

$$f''(x) = {e^x}(2) + (2x + 1){e^x}$$

$$ = {e^x}(2x + 3)$$

For $$x = - {1 \over 2}\,\,f''(x) > 0$$

$$\Rightarrow$$ Maxima

$$\therefore$$ Max. $$ = {e^{ - {1 \over 2}}}( - 1 - 1)$$

$$\therefore$$ $$ - {2 \over {\sqrt e }}$$