<s>[INST] Functional equation of function defined over non-negative reals satisfying $f \big(xf(y)\big)f(y)=f(x+y)$, $f(2)=0$ and $f(x)\ne0$ for $x\in[0, 2)$ 
Find all $ f:[0,\infty)\to [0,\infty) $ such that $ f (2)=0 $, $ f (x)\ne 0 $ for $ x\in [0, 2) $ and
$$ f \bigl(xf (y)\bigr) f (y)=f (x+y) $$
for all $ x, y\ge 0 $.

I tried plugging in values of $ x $ and $ y $ but didn't succeed.
Do you know how to proceed?
 [/INST] Substitute in $x=y=0$, we get that $ f(0) f(0) = f(0)$. Since $f(0) \neq 0$, thus $f(0) = 1$.
Substitute in $x=x, y = 2$, $0 = f(x\cdot 0) \cdot 0 = f(x+2)$. Hence for $x \geq 2, f(x) = 0 $.
We now focus our attention to the region $x\leq 2$.
Substitute in $x=2-y, y = y < 2$. We get that $ f[ (2-y) f(y) ] f(y) = f(2) = 0$. Since $f(y) \neq 0$ thus $(2-y) f(y) \geq 2$, or that $f(y) \geq \frac{2}{2-y}$.
Suppose that there exists a value $y$ such that $f(y) > \frac{2}{2-y}$. Then, take the value $x$ such that $ \frac{2}{f(y)} < x < 2-y$. We get that
$$ 0 = f( x f(y) ) f(y) = f(x+y) \neq  0,$$
which is a contradiction. Hence, at best, $f(y) = \frac{2}{2-y}$.
It is easy to check that $ f(x) = \begin{cases} \frac{2}{2-x}& 0\leq x < 2 \\ 0 & 2 \leq x \\ \end{cases}$ is a solution to the functional equation. (The simplest approach is to condition on $x+y<2$ and $x+y \geq 2$). Hence, it is the only solution.
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