Given: $f\left(0\right)=1,\underset{x\to -\infty }{\lim }f\left(x\right)=1$ 

And $f^{\text{'}}\left(x\right)-\alpha ·f\left(x\right)=3$

$\Rightarrow \frac{dy}{dx}-\alpha ·y=3$

Which is a linear differential equation, so

$\Rightarrow I.F.=e^{-\alpha x}$

Solution of the differential equation is given by,

$y\left(e^{-\alpha x}\right)=\int 3·e^{-\alpha x}dx$

$\Rightarrow f\left(x\right)\cdot \left(e^{-\alpha x}\right)=\frac{3e^{-\alpha x}}{-\alpha }+c$

$\Rightarrow f\left(x\right)=\frac{-3}{\alpha }+c·e^{\alpha x}$

Now, taking case I: when $\alpha >0$ and $\underset{x\to -\infty }{\lim }f\left(x\right)=1$ we get,

$\Rightarrow 1=\frac{-3}{\alpha }+c\left(0\right)$

$\alpha =-3$ (rejected)

Case-II: $\alpha <0$

as $\underset{x\to -\infty }{\lim }f\left(x\right)=1$

 $\Rightarrow c=0 \& \frac{-3}{\alpha }=1$ $\left\{\text{as} e^{-\left(-\infty \right)}\to \infty , \text{so }c=0\right\}$

$\Rightarrow \alpha =-3$

$\Rightarrow f\left(x\right)=1$ and also  $f\left(0\right)=1$

Hence, $f\left(x\right)=1$ is constant function, so $f\left(-{\log }_{e}2\right)=1$