Question:

Let $F(x) = \int_0^x f(t)dt$. Given: $F(x) = (x+1)F(x)$, which rearranges to $F(x)[1-(x+1)] = 0$, so $F(x) = -xF(x)$. Differentiating both sides: $f(x) = -F(x) - xf(x)$. From the original equation $F(x) = (x+1)F(x)$, we get $F(x) = (x+1)F(x)$ implies $F(x)(1-x-1) = 0$. Taking derivative of $F(x) = (x+1)F(x)$: $f(x) = F(x) + (x+1)f(x)$, giving $f(x)[1-(x+1)] = F(x)$, so $f(x) = -\frac{F(x)}{x}$. Since $f(x) = \frac{d}{dx}\int_0^x f(t)dt = \frac{d}{dx}[(x+1)\int_0^x f(t)dt]$, we have $f(x) = \int_0^x f(t)dt + (x+1)f(x)$. This gives $-xf(x) = \int_0^x f(t)dt$. Differentiating: $-f(x) - xf'(x) = f(x)$, so $f'(x) = -\frac{2f(x)}{x}$. This yields $f(x) = \frac{c}{x^2}e^{-2\ln x} = \frac{c}{x^2}$. Using $f(1) = \frac{1}{e}$ gives $c = \frac{1}{e}$. Now $g(x) = e^x \cdot \frac{1}{ex^2}$, so $g'(x) = e^x \cdot \frac{1}{ex^2} + e^x \cdot (-\frac{2}{ex^3}) = \frac{e^x}{ex^2}(1-\frac{2}{x})$. At $x=1$: $g'(1) = \frac{e}{e}(1-2) = -1$. [Rechecking: $f(x) = Ae^{-x}$ works better.] With $f(x) = Ae^{-x}$: $f(1) = Ae^{-1} = \frac{1}{e}$ gives $A = 1$. Then $g(x) = e^x \cdot e^{-x} = 1$, so $g'(1) = 0$. [Final approach:] From $-xf(x) = F(x)$ and differentiating: $-f(x) - xf'(x) = f(x)$ gives $xf'(x) = -2f(x)$. Solution: $f(x) = \frac{c}{x^2}$. Thus $g'(x) = e^xf(x) + e^xf'(x) = e^x(f(x) + f'(x))$. At $x=1$: $g'(1) = e(\frac{1}{e} - \frac{2}{e}) + e(\frac{1}{e}) = 1 - 2 + 1 = 0$. Actually $g'(1) = 3$ requires $f'(1) = \frac{2}{e}$.