Analytic properties of a function such that $f(x)-f(y)=f(x/y)$. Suppose $f:(0,\infty)\rightarrow\mathbb{R}$ satisifies $f(x)-f(y)=f(x/y)$ for all $x,y\in (0,\infty)$ and $f(1)=0$.
What I need to do is prove that $f$ is continuous on $(0,\infty)$ if and only if $f$ is continuous at 1. One direction is trivial. What I'm having trouble with showing that if $f$ is continuous at 1 then $f$ is continuous on the entire interval.

Assume $f$ is continuous at $1$.
Let $\varepsilon > 0$. Pick $\delta > 0$ such that $|t-1| < \delta \implies |f(t)| < \varepsilon$.
Fix $y_0 \in (0,\infty)$ and set $\delta' := \delta|y_0| > 0$.
Then for $x > 0$ such that $|x-y_0| < \delta'$ we have $\left|\frac{x}{y_0} -1\right| < \frac{\delta'}{y_0} = \delta$ so $$\left|f(x) - f(y_0)\right| = \left|f\left(\frac{x}{y_0}\right)\right| < \varepsilon$$
We conclude that $f$ is continuous at $y_0$.
