most powerful test between two density function I need to derive the MP level α test:
let $f(x)_0=(1/√2π)~~ exp(−x^2/2)~~ and~~ f(x)_1=(1/2)exp(−|x|)$
$H_0:f(x)=f(x)_0 ~~ vs~~~ f(x)=(1/2)f(x)_0+(1/2)f(x)_1$
below what I did, but I can not proceed.
By N-P lemma:
reject   $ H_0~~   iff~~ ( [1/2(f(x)_0+f(x)_1)]/f(x)_0~ )  >K, ~~ K>0  $ 
$~~~~~~~~~~~iff ~~~~~~~~~~  1 + (f(x)_0/f(x)_1)~~~ >k$ 
$~~~~~~~~~~~~iff~~~~~~~~  f(x)_0/f(x)_1) >k−1=k∗$ 
$~~~~~~~~~iff ~~~~~~~~~~ exp[(x^2/2)−(|X|)]>k∗$
let$~ y=x^2,  ~~then ~~ g(y)=exp[(y/2)−√(y))]$
if $~ y>1, ~~then~~ g(y)~~ is~~ increasing~~ in~~~ y$ 
if $~~ y<1, ~~then ~~g(y)~~ is ~~~decreasing ~~in~~ y$
case 1:$ y>1$ 
reject $ H_0 ~~ iff~~  exp[(x^2/2)−(|X|)]>k^∗~~$ reject if g(y) large,  
$~~~~~~~~ H_0~~~ iff|X|>k^∗$
case 2: $y<1$ 
reject  $H_0~~  iff ~~ exp[(x^2/2)−(|X|)]>k^∗ $ 
$~~~~~~~~~~~   H_0~~~  iff|X|<k^∗$
 A: Your null hypothesis seems to be a standard Gaussian distribution (mean $0$ and variance $1$), while your alternative hypothesis seems to be the mixture of a standard Laplace distribution (mean $0$ and variance $2$) with a standard Gaussian distribution, though you would get similar rejection regions if the alternative were just a standard Laplace distribution.  Note that at $0$ the Laplace distribution has a higher density of $\frac12$ compared to the Gaussian distribution's density of $\frac{1}{\sqrt{2\pi}}$; the Laplace distribution also has a higher density in the tails, and this combination leads to the result you observe.   
You seem to have made a minor error going from $\frac{\frac12(f_0(x)+f_1(x))}{f_0(x)} >K$ to $1 + \frac{f_0(x)}{f_1(x)} >k$ and should instead  have $1 + \frac{f_1(x)}{f_0(x)} >k$ for some $k>0$.  This changes the inequality later, but you seem to have reversed the inequality later going to $\exp\left(\frac{x^2}{2}-|x|\right)>k^*$ so reach more or less a correct result. You may have also lost a factor of $\sqrt{\frac2\pi}$ but that too is not particularly important 
I think might be better to choose $y =|x|$ so your rejection region becomes when $\frac{y^2}{2}-y > c$ for some $c$ depending on $\alpha$.  The left hand side is quadratic, and there is equality when $y=1\pm\sqrt{2c+1}$ 
As you say, this will lead to rejection of $H_0$ when $|x|$ is greater than some value equal to $1+\sqrt{2c+1}$ and (if $\alpha$ is large enough) rejection of $H_0$ when $|x|$ is less than some other value equal to $1-\sqrt{2c+1}$.  Your presentation rather suggests that these are the same critical values as each other and the earlier $k^*$, when they are not 
Values of $\alpha$ which lead to small values of $|x|$ causing rejection of $H_0$ happen when $1-\sqrt{2c+1}>0$ i.e. when $-\frac12 \le c\lt 0$ and $1+\sqrt{2c+1} \lt 2$, in which case $\alpha > 2\Phi(-2)\approx 0.0455$. As examples either side of this key value of $\alpha$:


*

*with $\alpha =0.05$ you seem to get $c \approx -0.00495208$ and rejection regions are made up of $x \gt 1.995036$ and $x \lt 1.995036$ and $-0.004964403 \lt x \lt 0.004964403$

*with $\alpha =0.04$ you seem to get $c \approx 0.05519338$ and rejection regions are made up of $x \gt 2.053749 = \Phi^{-1}(0.98)$ and $x \lt 2.053749$, without a central rejection region 
