Testing whether $X\sim\mathsf N(0,1)$ against the alternative that $f(x) =\frac{2}{\Gamma(1/4)}\text{exp}(−x^4)\text{ }I_{(-\infty,\infty)}(x)$ 
Consider the most powerful test of the null hypothesis that $X$ is a
  standard normal random variable against the alternative that $X$ is a
  random variable having pdf 
$$f(x) =\frac{2}{\Gamma(1/4)}\text{exp}(−x^4)\text{
 }I_{(-\infty,\infty)}(x)$$
and give the p-value if the observed value of $X$ is $0.6$

My try:
I think I should be using a likelihood ratio test.
I read that the Neyman–Pearson lemma states that the likelihood ratio test is the most powerful among all level $\alpha$  tests.
We have that the likelihood ratio is
$$\frac{f_0(x)}{f_1(x)}=\frac{\frac{1}{\sqrt{2\pi}}\text{exp}(-x^2/2)}{\frac{2}{\Gamma(1/4)}\text{exp}(-x^4)}=\frac{\Gamma(1/4)}{\sqrt{8\pi}}\text{exp}\left(\frac{-x^2}{2}+x^4\right)$$
Thus we accept $H_0$ if
$$\frac{\Gamma(1/4)}{\sqrt{8\pi}}\text{exp}\left(\frac{-x^2}{2}+x^4\right)\geq c$$
or equivalently if
$$\frac{-x^2}{2}+x^4 \geq \text{log}\left(\frac{\sqrt{8\pi}\cdot c}{\Gamma(1/4)}\right)$$
or equivalently if one of the following holds:
$$x^2\geq \frac{1+\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}$$
$$x^2\geq \frac{1-\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}$$
or equivalently if one of the following holds:
$$x\geq \sqrt{\frac{1+\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}}$$
$$x\leq -\sqrt{\frac{1+\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}}$$
$$x\geq \sqrt{\frac{1-\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}}$$
$$x\leq -\sqrt{\frac{1-\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}}$$
For a meaningful acceptance region we only consider the top two of the four constraints. Hence we reject if $$x\in\left(-\sqrt{\frac{1+\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}},\sqrt{\frac{1+\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}}\right)$$
We wish, under the null, that the probability that $X$ assumes a value in this range to be $0.05$
For this to occur, we need
$$\sqrt{\frac{1+\sqrt{1+16\cdot\text{log}(1.382746\cdot c)}}{4}}=0.06270678$$
But software gives that there are no solutions for $c\in\mathbb{R}$. Any suggestions or confirmation of my approach would be much appreciated.
 A: Possible way to proceed:
By N-P lemma, a most powerful test of size $\alpha$ for testing $H_0$ against $H_1$ is the indicator function
$$\varphi(x)=\mathbf1_{\lambda(x)>k}$$
, where $$\lambda(x)=\frac{f_{H_1}(x)}{f_{H_0}(x)}$$ and $k$ is so chosen that $$E_{H_0}\varphi(X)=\alpha$$
Now, 
$$\lambda(x)\propto\frac{e^{-x^4}}{e^{-x^2/2}}=\exp\left(-x^4+x^2/2\right)$$
Or, $$\ln \lambda(x)=\frac{x^2}{2}-x^4+\text{ constant }$$
Log being monotonically increasing, it suffices to study the nature of this function. You can consider this to be a function of $x^2$, plot the function, differentiate and check the sign of the derivative. We are interested to know whether $\lambda(x)$ is increasing or decreasing in $x$, or both increasing and decreasing for different values of $x$.
Ultimately you would have to reach some conclusion regarding $x$ from $\lambda(x)$ if possible, like 
$$\lambda(x)>k\implies k_1<x^2<k_2\implies c_1<x<c_2$$
Apart from possible relations between $c_1,c_2$ which can be found through the analysis above, additional constraint on $c_1,c_2$ is of course the size (or level) restriction.

The critical region gets complicated if you go into details.
For $\lambda(x)>k$, we have for some $k_1$,
\begin{align}
\frac{x^2}{2}-x^4&>k_1
\\\implies x^4-2x^2\cdot\frac{1}{4}+\frac{1}{16}&<\frac{1}{16}-k_1
\end{align}
Or, $$\left(x^2-\frac{1}{4}\right)^2<\frac{1}{16}-k_1$$
So for some $c$, the critical region is the set of all $x$ such that
\begin{align}
\left|x^2-\frac{1}{4}\right|<c \iff -c+\frac{1}{4}<x^2<c+\frac{1}{4}
\end{align}
You can stop here. If you want you can solve this further for $x$, but not necessary I think. 
If you can find $c$ such that $P_{H_0}(-c+\frac{1}{4}<X^2<c+\frac{1}{4})=\alpha$, then a level $\alpha$ MP test is ready using N-P lemma. You know that under $H_0$, $X^2$ has a $\chi^2_1$ distribution.
A: The test will reject $H_0$ for sufficiently large values of the ratio
$$\begin{align*}
\frac{2}{\Gamma\left(\frac{1}{4}\right)}\frac{\text{exp}\left(-x^4\right)}{\frac{1}{\sqrt{2\pi}}\text{exp}\left(-\frac{x^2}{2}\right)}
&=\frac{2\sqrt{2\pi}}{\Gamma\left(\frac{1}{4}\right)}\text{exp}\left(-x^4+\frac{1}{2}x^2\right)\\\\
&=\frac{2\sqrt{2\pi}}{\Gamma\left(\frac{1}{4}\right)}e^{\frac{1}{16}}\text{exp}\left(-x^4+\frac{1}{2}x^2-\frac{1}{16}\right)\\\\
&=c\cdot\text{exp}\left(-\left[x^2-\frac{1}{4}\right]^2\right)
\end{align*}$$
where $c$ is a positive constant. For the observed value of $x$, this ratio equals
$$c\cdot\text{exp}\left(-\left[0.6^2-\frac{1}{4}\right]^2\right)=c\cdot\text{exp}(-0.0121)$$
The desired p-value is the probability, when $H_0$ is true, that $X$ assumes any value $x$ such that $$c\cdot\text{exp}\left(-\left[x^2-\frac{1}{4}\right]^2\right)\geq c\cdot\text{exp}(-0.0121)$$
Such values of $x$ satisfy
$$\left[x^2-\frac{1}{4}\right]^2\leq 0.0121\Rightarrow\left|x^2-\frac{1}{4}\right|\leq0.11$$
Hence $$-0.11\leq x^2-\frac{1}{4}\leq0.11\Rightarrow 0.14\leq x^2 \leq 0.36$$
Then $$\sqrt{0.14}\leq x \leq 0.6 \text{ or } -0.6\leq x \leq -\sqrt{0.14}$$
Because the standard normal pdf is symmetric about $0$, the desired p-value is 
$$2\left[\Phi(0.6)-\Phi(\sqrt{0.14})\right]=0.16$$
