Draw a picture. I could not solve the problem without one.
There are two parts to the region where $f(x,y)\ne 0$, the first part in your list and the second part. The two parts are obtained from each other by interchanging the roles of $x$ and $y$, So there will be symmetry across the line $y=x$. The function $cxy$ we are integrating is also symmetric in $x$ and $y$, so the integrals over the two parts will be equal. Therefore we can just integrate over our favourite part, and double the result.
We look in detail at the second part of the region. It turns out that this is the triangle with corners $(0,0)$, $(0,-1)$, and $(1,0)$. At a certain point of the sketching, you will want to know on which side of the line $x-y=1$ our region lies. One way to decide is to rewrite the inequality $x-y \le 1$ as $y \ge x-1$, so $y$ is supposed to be bigger than $x-1$. But $y$ bigger means we are above the line $x-y=1$.
Express our integral over this part as an iterated integral, integrating first with respect to $y$, then with respect to $x$.
The "bottom" curve is $x-y=1$, the top curve is $y=0$. So when we integrate with respect to $y$, we integrate from $y=x-1$ to $y=0$. So we want $\int_{x=0}^1\left(\int_{y=x-1}^0 cxy\,dy\right)\,dx.$
The inner antiderivative is $\frac{cxy^2}{2}$. When we substitute our endpoints, we get $-\frac{cx(x-1)^2}{2}$. Now integrate from $x=0$ to $x=1$. To do the integration, you may want to expand out $x(x-1)^2$. And remember that this is half of the ultimate integral. This gives us an excuse to preemptively multiply by $2$.