Neyman-Pearson test at level $\alpha$ Let $P_0=\mathcal{N}(0,1)$ and $P_1=\mathcal{N}(\mu,\sigma^2)$ with $\sigma > 1$
I would like to show that the Neyman-Pearson test of $P_0$ vs. $P_1$ at level $\alpha$ has the form
$$\varphi_{*}(x)=\mathbb{1}_{[x \notin(\mu/(1-\sigma^2)\pm\delta_{*})]}$$
for some $\delta_{*}=\delta_{*}(\mu,\sigma,\alpha)>0$ and also to determine special case of $\varphi_{*}$ for $\mu=0$
What I have tried: Let $f_0$ be the density of probability distribution $P_0$, so $f_0=\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2})$ and similarly for $f_1=\frac{1}{\sqrt{2\pi} \sigma}\exp(-\frac{(x-\mu)^2}{2\sigma^2})$.
Then I found what is monotone density ratios:
$\frac{f_1}{f_0}=\frac{1}{\sigma} \frac{\exp(-\frac{(x-\mu)^2}{2\sigma^2})}{\exp(-\frac{x^2}{2})}=\frac{1}{\sigma} \exp(-\frac{(x-\mu)^2}{2\sigma^2}+\frac{x^2}{2})=g(T(x))$
$g(t)=\frac{1}{\sigma} \exp(-\frac{(t-\mu)^2}{2\sigma^2}+\frac{t^2}{2})$
and $T(x)=x$.
Then, let $H: \mathbb{R} \to [0,1]$ be an auxiliary function. 
$$H(r):=P_{0}(T>r)$$ for any $r \in \mathbb{R}$.
From this, we can determine $k_{\alpha}:=\min \{r \in \mathbb{R} : H(r) \leq \alpha\}$
and $$\gamma_{\alpha}=\frac{\alpha-P_0(T>k_{\alpha})}{P_0(T=k_{\alpha})} \in (0,1)$$
In my lecture notes I have $\varphi_{*}=\gamma_{\alpha}\mathbb{1}_{[T(x)=k_{\alpha}]}+\mathbb{1}_{[T(x)>k_{\alpha}]}$ for (UMP right sided-test) but I have problem with $\gamma_{\alpha}$ which in this case seems to be 1 and to define $k_{\alpha}$ and what about $\mathbb{1}_{[T(x)>k_{\alpha}]}$?
 A: I do not follow you well after you use the term $g(T(x))$ and introduce some auxiliary function. So I can not help you with that (maybe you can explain where that stuff is coming from).
But, this case has a more straightforward and simple solution starting from your expression for $\frac{f_1}{f_0}$

With the test you choose some $c$ such that
$$\frac{f_1}{f_0} = \frac{1}{\sigma} \exp \left( -\frac{(x-\mu)^2}{2\sigma^2} + \frac{x^2}{2}  \right)  >c$$
only occurs in a fraction of $\alpha$ cases when $P_0$ is true.

Form of equation
This can be converted by taking the logarithm
$$-\frac{(x-\mu)^2}{2\sigma^2} + \frac{x^2}{2}   >log (\sigma c)$$
and rewriting the quadratic equation:
$$ a_2 x^2  + a_1 x + a_0- log(\sigma c) > 0 $$
I let you do the work to get all of the coefficients but $a_2 = \frac{\sigma^2-1}{2\sigma^2}$ and $a_1 = \frac{\mu}{\sigma^2}$. Note that $a_2>0$ if  $\sigma>1$ so you have a parabola shape that opens upwards.
So you can imagine $\frac{f_1}{f_0}>c$ by the region outside of the region enclosed by the roots of the quadratic function:
$x_{_{f_1/f_0=c}} = \frac{-a_1}{2 a_2} \pm \frac{\sqrt{D}}{a_2} = \frac{-\mu}{\sigma^2-1} \pm \frac{\sqrt{D}}{a_2} =  \frac{\mu}{1-\sigma^2} \pm \delta_* $

Specific equation for $\mu=0$
You could work out the discriminant term $D$ and also express the level $c$ in terms of $\alpha$.
The discriminant is found when you work out the quadratic function in the section before.
The expression of $c$ in terms of $\alpha$ is found by expressing $P(\frac{f_1}{f_2}>c)=\alpha$.
But, note that now you have a simpler (symmetric) $$\frac{f_1}{f_0} = \frac{1}{\sigma} exp \left( \frac{\sigma^2-1}{2\sigma^2}   x^2 \right) $$
and
$$\varphi_{*}(x)=\mathbb{1}_{[x \notin(\pm\delta_{*})]} = \mathbb{1}_{[|x| >\delta_{*}]} $$
So you can look for $\delta_{*}$ such that $P(|x| >\delta_{*}) = \alpha)$ if $P_0$ is true. Which is an "ordinary" expression of the CDF of the standard normal.

More general
You might be able to do the same without much effort (if you allow computational solutions or expressions in terms of inverse, but not explicit, functions) for the more general case
$$\varphi_{*}(x)=\mathbb{1}_{[x \notin(\pm\delta_{*})]} = \mathbb{1}_{\left[ \left|x-\frac{\mu}{1-\sigma^2} \right| >\delta_{*}\right]} $$
so you are looking for an interval centered around $\frac{\mu}{1-\sigma^2}$ of size $2\delta_*$ that covers a fraction $1-\alpha$ of the standard normal distribution
