1. Given function

   $f(x)=e^{-|"log_e x|}, \, x\in(0,\infty).$

   Since $\log_e x=\ln x$, write

   $$$$

\ln x|}.$$

2. Remove the absolute value by cases

   We know:

   • If $x\ge 1$, then $\ln x\ge 0$, so $|\ln x|=\ln x$.
   • If $0<x<1$, then $\ln x<0$, so $|\ln x|=-\ln x$.

   Therefore,

   \begin{cases} e^{-(-\ln x)}=e^{\ln x}=x, & 0<x<1,\\[4pt] e^{-\ln x}=\dfrac{1}{x}, & x\ge 1. \end{cases}$$ So, $$f(x)= \begin{cases} x, & 0<x<1,\\ \dfrac{1}{x}, & x\ge 1. \end{cases}$$

3. Check continuity

   On $(0,1)$, $f(x)=x$, which is continuous.

   On $(1,\infty)$, $f(x)=\dfrac{1}{x}$, which is continuous.

   The only possible issue is at $x=1$.

   • Left limit: $\lim_{x\to 1^-} f(x)=\lim_{x\to 1^-} x=1.$
   • Right limit: $\lim_{x\to 1^+} f(x)=\lim_{x\to 1^+} \frac{1}{x}=1.$
   • Function value: $$$$

\ln 1|}=e^0=1.$$

Hence $f$ is continuous at $x=1$ as well.

Therefore, the number of points where $f$ is not continuous is $m=0.$

4. Check differentiability

   On $(0,1)$, $f(x)=x \implies f'(x)=1.$

   On $(1,\infty)$, $f(x)=\frac{1}{x} \implies f'(x)=-\frac{1}{x^2}.$

   Again, only $x=1$ needs checking.

   • Left derivative at $x=1$: $f'_-(1)=1.$
   • Right derivative at $x=1$: $f'_+(1)=-\frac{1}{1^2}=-1.$

   Since $f'_-(1)\ne f'_+(1),$ $f$ is not differentiable at $x=1$.

   Therefore, the number of points where $f$ is not differentiable is $n=1.$

5. Compute $m+n$

   $m+n=0+1=1.$

6. Match with options

   Option B: 1 is correct.