real analysis - I need to find all functions $f:mathbb R rightarrow mathbb R$ which are continuous and satisfy $f(x+y)=f(x)+f(y)$

I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $f(x+y)=f(x)+f(y)$. I know that there are other questions that are asking the same thing, but I'm trying to figure this out by myself as best as possible. Here is how I started out:

Try out some cases:

$x=0:$
$$f(0+y)=f(0)+f(y) \iff f(y)=f(0)+f(y) \iff 0=f(0) $$
The same result is for when $y=0$

$x=-y:$

$$f(-y+y)=f(-y)+f(y) \iff f(0)=f(-y)+f(y) \iff 0=f(-y)+f(y)\iff \quad f(-y)=-f(y)$$
I want to extend the result of setting $x=-y$ to numbers other that $-1$, perhaps all real numbers or all rational numbers. I got a little help from reading other solutions on the next part:

Let $q=1+1+1+...+1$. Then
$$f(qx)=f((1+1+...+1)x)=f(x+x+...+x)=f(x)+f(x)+...+f(x)=qf(x)$$
I understood this part, but I don't understand why this helps me find all the functions that satisfy the requirement that $f(x+y)=f(x)+f(y)$, but here is how I went on:

Thus
$$f(qx)=qf(x)$$ and it should follow that
$$f \bigg (\frac {1}{q} x\bigg)= \frac{1}{q}f(x)$$ where $q\not =0$, then it further follows that

$$f \bigg (\frac {p}{q} x\bigg)= \frac{p}{q}f(x)$$ where $\frac{p}{q}$ is rational, and lastly it further follows that
$$f (ax)= af(x)$$ where $a$ is real. Thus functions of the form $f(ax)$ where $a$ is real satisfies the requirement of $f(x+y)=f(x)+f(y)$.

I don't know how much of what I did is correct\incorrect, and any help would be greatly appreciated. Also is there any way that I can say that functions of the form $f(ax)$ where $a$ is real are the only functions that satisfy the requirement of $f(x+y)=f(x)+f(y)$? Or do other solutions exist?

Again, thanks a lot for any help! (Hints would be appreciated, I'll really try to understand the hints!)