Split the numerator into two pieces and take the limit term-by-term.

Piece 1: $\frac{e^x(e^{\tan x-x}-1)}{\tan x-x}=e^x\cdot \frac{e^{\tan x-x}-1}{\tan x-x}$.

As $x\to 0$: $e^x\to 1$ and $\frac{e^u-1}{u}\to 1$ where $u=\tan x-x\to 0$. Limit $=1$.

Piece 2: $\frac{{\log }_e(\sec x+\tan x)-x}{\tan x-x}$.

Using the series ${\log }_e(\sec x+\tan x)=x+\frac{x^3}{6}+O(x^5)$ (the antiderivative of $\sec x$) and $\tan x-x=\frac{x^3}{3}+O(x^5)$:

Limit $=\frac{x^3/6}{x^3/3}=\frac{1}{2}$.

Therefore $f(0)=1+\frac{1}{2}=\frac{3}{2}$.