If we have the function:
$$f_{X,Y}\left(x,y\right)=\frac{3}{7}x,\:\text{ when }\:1\le x\le 2\:\wedge \:0\le y\le x$$
If I want to find the marginal function $f_Y\left(y\right)$.
It should be: $$f_Y\left(y\right)=\int _1^2\:\frac{3x}{7}\,dx=\frac{9}{14}.$$
Is it right or I need to split something?
You can always check by integrating (it should give you $1$, but in your case it doesn't give $1$ after integrating)
Your density is given by $$ f_{X,Y}(x,y) = \begin{cases} \frac{3}{7}x & x \in [1,2], y \in [0,x] \\ 0 & otherwise \end{cases} $$ To find marginal distribution of $Y$, you need to fix $y$ in support of your density, that is $y \in [0,2]$ and then integrate (with that fixed $y$) with respect to $x$ over whole space, that is you need to integrate:
$$ f_{Y}(y) = \int_{\mathbb R}f_{X,Y}(x,y)dx $$ Note that when $y \in [0,1)$ then $x$ can take any value between $[1,2]$ hence for $y \in [0,1)$ you indeed have $$f_{Y}(y) = \int_{1}^2 \frac{3}{7}x dx = \frac{3}{14}(2^2-1^2) = \frac{9}{14} $$ But when $y \in [1,2]$, then note that you have requirement $y \in [0,x]$, hence $x$ is not supported by whole $[1,2]$ but only $[y,2]$ (and that is okay that support of $x$ depends on $y$, our density can depend on $y$). So in case $y \in [1,2]$ you'd rather have $$ f_{Y}(y) = \int_{y}^2 \frac{3}{7}x dx = \frac{3}{14}(2^2 - y^2) = \frac{6}{7} - \frac{3y^2}{14} $$
So that marginal density is given by $$ f_{Y}(y) = \begin{cases} \frac{9}{14} & y \in [0,1) \\ \frac{6}{7} - \frac{3y^2}{14} & y \in [1,2] \\ 0 & otherwise \end{cases} $$