Let $X$ have probability density of $f_X=xe^{-x}dx$ on $\mathbb{R}^+$ and $U$ the uniform distribution on $[0,1]$ independent of $X$. Define $Y_1=UX$ and $Y_2=(1-U)X$. The goal is now to show that $Y_1$ and $Y_2$ are independent. The proof I'm looking at works as follows:
First, considering the distribution of $(Y_1,Y_2)$ using expected value $E(f(UX,(1-U)X))=\int_0^1 \int_0^\infty xe^{-x}f(ux,f((1-u)x))dxdu = \int_0^\infty \int_0^\infty f(s,t)e^{-(t+s)}dsdt$.
Then, the proof says that since we can represent the law of $(Y_1,Y_2)$ as product of the law of $Y_1$ and $Y_2$, they're independent.
But how can we do this? How can we know that $Y_1$ and $Y_2$ have laws $e^{-t}dt$, $e^{-s}ds$ respectively?
$Y_1$ and $Y_2$ are jointly distributed as independent exponentials. The easiest way to see this is probably to just do a change of variables. We have $$ X = Y_1+Y_2\\U = \frac{Y_1}{Y_1+Y_2},$$ giving a Jacobian of $\frac{1}{y_1+y_2}.$ Then we have $$ f_{Y_1,Y_2}(y_1,y_2) = \frac{1}{y_1+y_2}f_{X,U}\left(y_1+y_2, \frac{y_1}{y_1+y_2}\right) = \frac{1}{y_1+y_2}(y_1+y_2) e^{-(y_1+y_2)}= e^{-(y_1+y_2)}.$$
This solves the problem, but I assume the theorem you're trying to use is that $Y_1$ and $Y_2$ are independent if and only if $$ E(g_1(Y_1)g_2(Y_2)) = E(g_1(Y_1))E(g_2(Y_2))$$ for all (sufficiently well behaved) functions $g_1$ and $g_2.$ So to apply your integral to this, you need to take $f(x,y) = g_1(x)g_2(y),$ and you have $$ E(g_1(Y_1)g_2(Y_2)) \\= \int_0^\infty \int_0^\infty g_1(s)g_2(t) e^{-(s+t)}dsdt \\= \left(\int_0^\infty g_1(s) e^{-s}ds\right)\left(\int_0^\infty g_2(t) e^{-t}dt\right) \\= E(g_1(Y_1))E(g_2(Y_2))$$ since $Y_1$ and $Y_2$ are both exponentially distributed. Of course using our previous results is overkill here since they already show independence. However, perhaps the intention was to show that $Y_1$ and $Y_2$ are marginally exponential, through some other method, for instance you can do the integral $$ P(Y_1\le y ) = P(X\le y/U) = \int_0^1 du\int_0^{y/u} xe^{-x} dx = 1-e^{-y},$$ which works out, but I find no less burdensome than the change of variables above.