I just want to have certain that I'm thinking right
If $$f(x,y)=\frac{12}{5}\times x\times (2-x-y)$$for $0<x<1, 0<y<1$ and $f(x,y)=0$ c.c. the conditional distribution of X if $0<y<\frac{1}{2}$. $$f_{X|Y}(x|y)=\frac{f(x,y)}{f_Y(y)}=\frac{\frac{12}{5}\times x\times (2-x-y)}{\int_{0}^{\frac{1}{2}}\frac{12}{5}\times x\times (2-x-y)dy}$$
I think the conditional distribution of X, if $0<y<\frac{1}{2}\text{ should be something like below:}$ $$f_{X|Y}(x|y)=\frac{f(x,y)}{f_Y(y)}=\frac{\frac{12}{5}\times x\times (2-x-y)}{\int_{y=0}^{y=\frac{1}{2}}\int_{x=0}^{{x=1}}\frac{12}{5}\times x\times (2-x-y)dxdy}$$