Given a continuous random vector has density $f_{X,Y}(x,y) = 2(x + y) :0 \lt x \lt y \lt 1.$
Find $\mathbb E(XY)$.
I'm unclear on whether the method would be to perform the integral: $$\int_0^1 \int_0^yxyf_{X,Y}(x,y)dxdy$$ by using the law of the unconscious statistician or whether there is another approach to this.
Your way is the best approach. An unattractive alternative is to let $W=XY$ and find the density function of $W$, then calculate $E(W)$ in the usual way.
To find the density function of $W$, we can for example calculate the cumulative distribution function of $W$ and then differentiate.
For the cdf of $W$, we find the probability that $W\le w$. For a $w$ between $0$ and $1$, draw the curve $xy=w$. Let $K_w$ be the part of the triangle with corners $(0,0)$, $(1,1)$, and $(0,1)$ that is "below" $xy=w$. Then $\Pr(W\le w)$ is the integral over $K_w$ of the joint density of $X$ and $Y$.
There are other ways to find the density function of $W$.