$$f(y|x)= \begin{cases} \frac{1}{x},& \text{if } 0\leq y \leq x\\ 0, & \text{otherwise} \end{cases}$$
$$f(x)= \begin{cases} 2x,& \text{if } 0\leq x \leq 1\\ 0, & \text{otherwise} \end{cases}$$
I need to find $f(x,y)$ but I don’t know how to multiply the conditional function $f(y|x)*f(x) = f(x,y)$
What you can observe, is that if not $0\le y\le x\le 1$, then one of the two functions will be zero. So we have: $$ f(y|x)\cdot f(x) = \begin{cases} 2 & 0\le y \le x \le 1 \\ 0 & \textrm{otherwise} \end{cases} $$ The domain is the triangle bounded by the lines $y=0$, $x=1$ and $y=x$. To be very explicit: $$ f(y|x)\cdot f(x) = \begin{cases} \frac1x \cdot 2x & y\in[0,x] \wedge x\in[0,1] \\ 0 \cdot 2x & y\notin[0,x] \wedge x\in[0,1] \\ \frac1x \cdot 0 & y\in[0,x] \wedge x\notin[0,1] \\ 0\cdot0 & y\notin[0,x] \wedge x\notin[0,1] \\ \end{cases} $$ Makes sense?