Given that, $f(x)=\frac{x-1}{e^x}$ $\ldots(\mathrm{i})$
Differentiating Eq. (i) w.r.t. $x$ on both sides, we get
$f^{\prime}(x)=\frac{e^x \frac{d}{d x}(x-1)-(x-1) \frac{d}{d x}\left(e^x\right)}{\left(e^x\right)^2}$
$\left[\because \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d}{d x} u-u \frac{d v}{d x}}{v^2}\right]$
$=\frac{e^x \cdot 1-(x-1) e^x}{e^{2 x}}$
$f^{\prime}(x)=\frac{2-x}{e^x}$ $\ldots(\mathrm{ii})$
Putting $x=0$ in Eq. (ii), we get
$f^{\prime}(0)=\frac{2-0}{e^0}=\frac{2}{1} \Rightarrow f^{\prime}(0)=2$ $\ldots(\mathrm{iii})$
Now, differentiating Eq. (ii) w.r.t. $x$ on both sides, we get
$f^{\prime \prime}(x)=\frac{e^x \frac{d}{d x}(2-x)-(2-x) \frac{d}{d x}\left(e^x\right)}{\left(e^x\right)^2}$
$=\frac{e^x(-1)-(2-x) e^x}{e^{2 x}}$
$f^{\prime \prime}(x)=\frac{(x-3)}{e^x}$ $\ldots(\mathrm{iv})$
Putting $x=0$, in Eq. (iv), we get
$f^{\prime \prime}(0)=\frac{0-3}{e^0}=\frac{-3}{1} \Rightarrow f^{\prime \prime}(0)=-3$ $\ldots(\mathrm{v})$
Adding Eq. (iii) and Eq. (v), we get
$f^{\prime}(0)+f^{\prime \prime}(0)=2+(-3) \Rightarrow f^{\prime}(0)+f^{\prime \prime}(0)=-1$