Suppose $f(t) = 0$ for some $t > 0$. For $0 < x < t$ we choose $y = t - x > 0$ and find $f(x) = 0$. From setting $x = y = 1$ we conclude that $f(1) \ne -1$. Hence by setting $x = y$ we get $f(x) - f(2x) = f(1)f(2x)$ for $x > 0$. Inductively we find
$$f(2^n x) = f(x)(1 + f(1))^{-n}. \quad (2)$$
Hence for an arbitrary $x > 0$ we choose $n \in \mathbb{N}$ such that $\frac{x}{2^n} \le t$ and we conclude from (2) that $f(x) = 0$.

Now suppose $f(t) \ne 0$ for all $t > 0$. We define $g(x) := f(x)^{-1}$, $x > 0$, and rewrite the given equation as
$$g(x + y) - g(x) = \frac{g(x)}{g\left(\frac{x}{y}\right)} \quad \text{for all } x, y > 0. \quad (3)$$
By setting $y = 1$ we obtain $g(x + 1) = g(x) + 1$ for all $x > 0$. It follows that $g(n) = n - 1 + g(1)$ holds for all $n \in \mathbb{N}$. Setting $x = y = 2$ in (3) we now find $g(1) = 1$. Setting $x = 1$ we obtain $g(y)g(\frac{1}{y}) = 1$ for all $y > 0$. We can now rewrite (3) as
$$g(x + y) = g(x) + g(x)g\left(\frac{y}{x}\right) = g(x)\left(1 + g\left(\frac{y}{x}\right)\right) = g(x)g\left(\frac{x+y}{x}\right). \quad (4)$$
From (4) we directly see that $g(a)g(b) = g(ab)$ for all $a > 0$ and $b > 1$. Using $g(y)g(\frac{1}{y}) = 1$, we can rewrite (4) as
$$g\left(\frac{x}{x+y}\right)g(x + y) = g(x),$$
which together with the previous line shows that $g(a)g(b) = g(ab)$ holds indeed for all $a, b > 0$. In particular we see that $g(a) = g(\sqrt{a})^2 > 0$, so $g$ is a positive function. Also, from the first equation in (4) we now infer the functional equation
$$g(x + y) = g(x) + g(y) \quad \text{for all } x, y \in (0, \infty).$$
It is well known that this implies $g(x) = xg(1) = x$ for $x \in \mathbb{Q}_{>0}$. Since $g$ is positive, by (3) we deduce that $g(x + y) > g(x)$, so $g$ is strictly increasing. Hence $g(x) = x$ for all $x > 0$.

Since $f(x) = 0$ and $f(x) = \frac{1}{x}$ obviously satisfy the given equation, we have found all solutions. $\square$