Does a function $f(x,y)$ exist that $\int_0^{1-x}f(x,y)dy=1$ and $\int_0^{1-y}f(x,y)dx=1$? Suppose function $f(x,y)$ is defined on triangle $x\ge 0, y\ge 0, x+y\le 1$. Does such a function exist that $\int_0^{1-x}f(x,y)dy=1$ and $\int_0^{1-y}f(x,y)dx=1$ for any $x\in (0,1)$ and $y\in (0,1)$?
Actually, I'm trying to find a probability distribution that both marginal distributions are uniform.
 A: I shall answer not the question of your title, but the one behind (finding some probability distribution). Let $$T := \{x \geq 0\} \cap \{y \geq 0\} \cap \{x+y \leq 1\}$$
I claim that there is only one distribution on $T$ such that both marginals are uniform on $[0,1]$. This distribution is uniform on the side $L = \{x+y = 1\} \cap T$. In particular, this distribution is not absolutely continuous with respect to the Lebesgue measure (it has no density with respect to the Lebesgue measure).
Let $\mu$ be such a distribution. Note that $x$ and $y$ both are uniform on $[0,1]$. Hence, $\mathbb{E}_\mu (x+y) = \mathbb{E}_\mu (x)+\mathbb{E}_\mu (y) = 1$.
Assume that $\mu(L) < 1$. Then $\mu (\{x+y < 1\}) > 0$. By $\sigma$-additivity, there is $\varepsilon > 0$ such that $\delta := \mu (\{x+y < 1- \varepsilon\}) > 0$. But then $\mathbb{E} (x+y) \leq  \delta(1-\varepsilon) + (1-\delta) = 1-\delta \varepsilon < 1$, and we have a contradiction.
Hence, $\mu(L) = 1$. Then $\mu (\{(t,1-t): t \in (a,b)\}) = b-a$ whenever $(a,b) \subset [0,1]$, as its marginals are uniform. This characterizes $\mu$.
This does not means that your equations have no solution; however, if a solution exists, it cannot be nonnegative (or perhaps measurable).
