Given $X \sim u(0,1)$, we define $Y=1-X$, then we have that $f_{X}(x)=I_{[0,1]}(x)$ and $F_{X}(x)=xI_{[0,1]}(x) + I_{(1, \infty)}(x)$. I know, if $0\le y \le 1$
$$F_{Y}(y)=P[Y \le y]=P[1-X \le y]=P[1-y \le X]=1-P[X <1-y]=1-F_{X}(1-y)=1-(1-y)=y$$
therefore $Y \sim u(0,1)$ and $f_{Y}(y)=I_{[0,1]}(y)$.
Because $X,Y$ are not independent, how can I find $f_{X,Y}(x,y)?$ I will receive any idea, thanks in advance.
Indeed $X$ and $Y$ are both uniform on $(0,1)$. Let $D=\{(x,y)\mid x\gt0,y\gt0,x+y=1\}$, then the set $D$ has zero Lebesgue measure and $P[(X,Y)\in D]=1$ hence $(X,Y)$ has no density.
A way to characterize the resulting distribution in such cases is to provide the value of $E[u(X,Y)]$ for every (measurable bounded) function $u$. Here, $$ E[u(X,Y)]=\int_0^1u(x,1-x)\,\mathrm dx. $$