We know that the common pdf of $X,Y$ is constant function, on the triangle $(0,0),(0,1),(2,0)$ (and out of this range the value of the function is zero).
What is $f_X(x)$ and $f_Y(y)$?
My solution:
The area of the triangle is 1, so the constant function is:
$$f_{XY}(x,y)=\cases{1\;\;0<x<2,0<y<1\\0\;Else}\\ $$
$1=\frac{1}{\text{the area of the triangle}}=\frac11=1$
So:
$$f_X(x)=\int _{-\infty}^{\infty}f_{XY}(x,y)dy=\int _{0}^{-\frac{x}{2}+1}1\,dy=-\frac{x}{2}+1\\f_Y(y)=\int _{-\infty}^{\infty}f_{XY}(x,y)dx=\int _{0}^{2-2y}1\,dy=2-2y$$
I'm right?
Or I miss something?
Thank you!
Hint:
$f_{X,Y}$ is prescribed by $\left(x,y\right)\mapsto1$ if $\left(x,y\right)\in\triangle$ and $\left(x,y\right)\mapsto0$ otherwise, where $\triangle$ denotes the triangle.
Then $f_{X}\left(x\right)=\int f_{X,Y}\left(x,y\right)dy=\int_{\triangle\left(x\right)}dy=\lambda\left(\triangle\left(x\right)\right)$ where $\triangle\left(x\right)=\left\{ y\mid\left(x,y\right)\in\triangle\right\} $ and $\lambda$ denotes the Lebesguemeasure.
Addendum:
Here $\triangle:=\left\{ \left(x,y\right)\in\left(0,2\right)\times\left(0,1\right)\mid x+2y<2\right\} $. Note that that the lines $2y+x=2$, $x=0$ and $y=0$ are the 'borders' of the triangle.
Then for $x\in\left(0,2\right)$ you have $\Delta\left(x\right)=\left\{ y\in\left(0,1\right)\mid y<1-\frac{1}{2}x\right\} =\left(0,1-\frac{1}{2}x\right)$. Consequently for $x\in\left(0,2\right)$ we find $f_{X}(x)=\lambda\left(\left(0,1-\frac{1}{2}x\right)\right)=1-\frac{1}{2}x$.
If $x\notin\left(0,2\right)$ then $\Delta\left(x\right)=\emptyset$ leading to $f_{X}\left(x\right)=0$.
Likewise you can define $\Delta'\left(y\right)=\left\{ x\mid\left(x,y\right)\in\triangle\right\} =\left\{ x\in\left(0,2\right)\mid x<2-2y\right\} $ leading to $f_{Y}\left(y\right)=2-2y$ for $y\in\left(0,1\right)$ and $f_Y(y)=0$ if $y\notin(0,1)$.