Hypergeometric function and double integral Mathematica shows that
$$
\int _0^{\infty }\int _y^{\infty }e^{-\frac{x^3}{3}-\frac{5 y^3}{6}}dxdy = \frac{2 \pi  \, _2F_1\left(\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{5}{2}\right)}{3^{5/6} \Gamma \left(\frac{1}{3}\right)}.
$$
But it does not give any clue of how to prove this.
See WolframAlpha.
I have tried to use this representation
$$
 _2F_1(a,b;c;z)=\frac{1}{\Gamma\left(b\right)\Gamma\left(c-b\right)}\int_{0}^{1}\frac{t^{b-1}(%
1-t)^{c-b-1}}{(1-zt)^{a}}\mathrm{d}t.
$$
In the end, I could show that what I want to prove it is equivalent to 
$$
B\left(\frac{2}{3},\frac{2}{3}\right) \int_0^1 \frac{t^{\frac{1}{3}-1}}{\left(\frac{5 t}{3}+1\right)^{2/3}} \, dt
=
\int_0^1 \frac{3 t^{\frac{2}{3}-1}}{\sqrt[3]{1-t} \sqrt[3]{\frac{5 t}{3}+1}} \, dt.
$$
Then I got stuck here and could not go further.
 A: According to Gradshteyn-Ryzhik (3.194.1 in edition 6), (in fact, they use $u=t^3$)

$$\frac{d}{dt} t\,{}_2F_1(2/3,1/3,4/3,-t^3)=\frac{1}{(1+t^3)^{2/3}}.$$

With that in hand, the path is not too strange.
Let me give some more details (although, you might already know them). I suggest that you switch to polar coordinates. You will get
$$
\int_0^{\pi/4}\int_0^{+\infty}e^{-(\cos(\theta)^3/3+5\sin(\theta)^3/6)r^3}r\,dr\,d\theta.
$$
Integrating in $r$, you will get an incomplete gammafunction, and if I insert the limits correctly, the result is
$$
\int_0^{\pi/4}\frac{\Gamma(2/3)}{3\bigl(\cos(\theta)^3/3+5\sin(\theta)^3/6\bigr)^{2/3}}\,d\theta.
$$
Next, with $t=\tan\theta$ we get
$$
\begin{align}
\int_0^1\frac{\Gamma(2/3)}{3(1/3+5t^3/6)^{2/3}}\,dt&=\Bigl[\frac{t\Gamma(2/3)}{3^{1/3}}{}_2F_1(2/3,1/3,4/3,-5t^3/2)\Bigr]_0^1\\
&=\frac{\Gamma(2/3)}{3^{1/3}}{}_2F_1(2/3,1/3,4/3,-5/2).
\end{align}
$$
Note that the arguments $1/3$ and $2/3$ inside the Hypergeometric function can change place, and also that the reflection formula
$$
\Gamma(2/3)\Gamma(1/3)=\frac{2\pi}{\sqrt{3}}
$$
can be used to force the form to be the same as yours.
