How to evaluate $\int_{0}^{2\pi}dx\!\int_{0}^{\pi}\sin ye^{\sin y(\cos x-\sin x)}dy=\sqrt{2}\big(e^{\sqrt{2}}-e^{-\sqrt{2}}\big)\pi $

I will evaluate this integral indirectly by recognizing it for what it is: an integral over the surface of a unit sphere.

First of all, for simplicity sake, I am going to replace your notation so it looks like we are integrating over solid angle:

$$\int_0^{2 \pi} d\phi \, \int_0^{\pi} d\theta \, \sin{\theta} \, e^{\sin{\theta} (\cos{\phi}-\sin{\phi})} $$

Now, note that we can replace the $\cos{\phi}-\sin{\phi}$ in the exponential with simply $\sqrt{2} \cos{\phi}$, as we are integrating over the whole azimuth anyway. Thus, the integral is

$$\int_0^{2 \pi} d\phi \, \int_0^{\pi} d\theta \, \sin{\theta} \, e^{\sqrt{2} \sin{\theta} \cos{\phi}} $$

Now, note that on the unit sphere, $\sin{\theta} \cos{\phi} = x$, the $x$-coordinate. Thus, by simple geometry, we can replace this double integral with a single integral over $x$ by noting that the function we are integrating over only depends on $x$. The contribution to the integral from a ring a distance $x$ from the plane $x=0$ is $2 \pi \sqrt{1-x^2} e^{\sqrt{2} x} ds$, where $ds$ is an element of arc length. Of course, over a cross-section of the sphere, $y=\pm \sqrt{1-x^2}$ and

$$ds = \sqrt{1+\left ( \frac{dy}{dx} \right )^2} dx = \frac{dx}{\sqrt{1-x^2}}$$

Thus the integral is simply

$$2 \pi \int_{-1}^1 dx \, e^{\sqrt{2} x} $$

which produces the sought-after result.

This answer complements the solution presented by Ron Gordon. Note that we have

$$\begin{align} I&=\int_0^{2\pi}\int_0^\pi \sin(\theta)e^{\sin(\theta)(\cos(\phi)-\sin(\phi))}\,d\theta\,d\phi \tag 1\\\\ &=\int_0^{2\pi}\int_0^\pi \sin(\theta)e^{\sqrt{2}\sin(\theta)\cos(\phi)}\,d\theta\,d\phi \tag 2\\\\ &=\oint_{|r|=1}e^{\sqrt{2}x}\,dS \tag 3\\\\ &=2\int_{-1}^1 e^{\sqrt{2}x}\left(\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac{1}{\sqrt{1-x^2-y^2}}\,dy\right)\,dx \tag 4\\\\ &=2\int_{-1}^1 e^{\sqrt{2}x}\left.\left(\arctan\left(\frac{y}{\sqrt{1-x^2-y^2}}\right)\right|_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\right)\,dx \tag 5\\\\ &=2\int_{-1}^1 e^{\sqrt{2}x}\left(\pi\right)\,dx \tag 6\\\\ &=2\pi\int_{-1}^1 e^{\sqrt{2}x}\,dx \tag 7 \end{align}$$

In going from $(1)$ to $(2)$, we used the identity $\cos(\phi)-\sin(phi)=\sqrt{2}\cos(\phi+\pi/4)$ and exploited the $2\pi$-periodicity of the integrand with respect to $\phi$.

In going from $(2)$ to $(3)$, we noted that the integral is the closed-surface integral over the unit sphere.

In arriving at $(4)$, we exploited symmetry of the integration over upper and lower hemispheres, translated to Cartesian coordinates, and used $dS =\frac{1}{\sqrt{1-x^2-y^2}}\,dx\,dy$.

In going from $(4)$ to $(5)$, we began carrying out the inner integral on $y$.

In arriving at $(6)$, we completed the evaluation of the inner integral on $y$, which yielded $\pi$.

And finally in $(7)$, we simply factor the $\pi$ from the inner integral.