The question requires me to show that the large eigenvalues of the system $$\frac{d^2w}{dx^2}=(x^4+x^2-\lambda^2)w$$ $w(\infty)=w(-\infty)=0$
$$\boxed{\lambda\sim 2^{\frac{1}{3}}\pi\left(\Gamma\left(\frac{1}{4}\right)\right)^{-\frac{4}{3}}\left(3n+\frac{3}{2}\right)^{\frac{2}{3}}+O(n^\frac{1}{3})}$$
Using the WKB approximation theory of matching I find the following Bohr–Sommerfeld quantization condition:
$$\int_{-\alpha}^{\alpha}\sqrt{\lambda^2-x^2-x^4}dx=(n+1/2)\pi$$
Where $\displaystyle \alpha=\sqrt{\frac{-1+\sqrt{1+4\lambda^2}}{2}}$
I'm not really sure how to simplify this to give the above.
Not a full solution, but key steps.
The energy ($E=\lambda^2$) you search is defined at the stopping point ($x=\alpha$): $E=\alpha^4+\alpha^2$
You are using the semiclassical approximation and $\lambda^2>>1$ in your notations. Thus, we can drop $x^2$ and work only with $x^4$ to get the main asymptotic term: $E\approx\alpha^4$
Now, let's consider Bohr–Sommerfeld quantization conditions: $$I(\alpha)=\int_{-\alpha}^{\alpha}\sqrt{\lambda^2-x^2-x^4}dx\approx\int_{-\alpha}^{\alpha}\sqrt{\alpha^4-x^4}dx$$$$=\alpha^3\int_{-1}^{1}\sqrt{1-t^4}dt=\frac{\alpha^3}{2}\int_0^1\sqrt{1-u}u^{-\frac{3}{4}}du$$ $$I(\alpha)\approx\frac{\alpha^3}{2}B\,(\frac{3}{2};\frac{1}{4})=\frac{\alpha^3}{2}\frac{\Gamma(\frac{3}{2})\Gamma(\frac{1}{4})}{\Gamma(\frac{7}{4})}=\frac{\alpha^3}{3}\frac{\sqrt{\pi}\,\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}=\frac{\alpha^3}{3}\frac{\sin\frac{\pi}{4}\,\Gamma\,^2(\frac{1}{4})}{\sqrt{\pi}}$$
At the given level of approximation $E\approx\alpha^4\approx\lambda^2$, and $\alpha$ is defined by$$I(\alpha)\approx\frac{\alpha^3}{3}\frac{\sin\frac{\pi}{4}\,\Gamma\,^2(\frac{1}{4})}{\sqrt{\pi}}\approx(n+1/2)\pi$$
Please check the calculations.
For the reference I would recommend "Mathematical Methods for Physics and Engineering" of K.F. RILEY, M.P. HOBSON and S. J. BENCE. You may look through the section 25.7 (WKB method), pp 900-902, where the similar example is reviewed in details.