Prove that there is some constant $a\in \mathbb R$ such that $f(ax)=af(x)$ for every $x \in \mathbb R$

Question

My attempt:-

I could prove that $f(x_0)=f(1)x_0.$ Using the continuity of $x_0$. How can I prove this for every $x\in \mathbb R$? It is given that $f$ is continuous only at some $x_0 \in \mathbb R.$ If I knew the countinuity of $f$ in entire $\mathbb R$, I could prove it similarly. Please help me.

Answer 1

$|f(y)-f(x)|=|f(y-x)|=|f(y-x+x_0)-f(x_0)| \to 0$ as $y \to x$ because $y-x+x_0 \to x_0$. So $f$ is continuous at any point $x$.