For positive integer $n$ we have $$f(n) = \frac{n}{n+1}f(n-1) = \frac{n-1}{n+1}f(n-2) = \cdots = 0\times f(-1) = 0$$ Hence $f(x)$ has infinitely many zeros, and must be identically zero, i.e. $f(x)=0$.