= Solution
Use $u=x-y$, $v=x+y$. It sends the diamond-shaped open square onto $(-1,1)^2$, and the inverse Jacobian has absolute <determinant> $1/2$. On this rectangle $|u|<1$ and $v+2>1$, so the original function is measurable, bounded by one and supported on a finite-area set, proving integrability. The transformed integral separates:
$$
\int f^{2n}=\frac12\int_{-1}^1u^{2n}\,du\int_{-1}^1(v+2)^{-2n}\,dv
=\boxed{\frac{1-3^{1-2n}}{(2n+1)(2n-1)}}.
$$
The two one-dimensional integrals follow from the <fundamental theorem of calculus>; Fubini's theorem applies by the established integrability.
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