I was studying the functional equation $f\bigl(xf(y)\bigr)f(y)=f(x+y)$ and observed that for $x=\delta y$ we can write $f(y+\delta y)-f(y) = f(y)\Bigl(f\bigl(\delta yf(y)\bigr)-1\Bigr)$. I then reasoned that, as $f(0)=1$ we can make $f(y+\delta y)-f(y)$ arbitrarily small by decreasing $\delta y$ so the function ought to be continuous. I want to know whether this proves the function is continuous or the proof requires $f$ to be continuous at $x=0$ as well. Also what can I do to prove this (apart from solving the equation)?
The solution, by the way, is $f(x)= \frac{2}{2-x}$ for $x\ne2$.
If I understand you right you're saying :
$f(y+\delta y)-f(y) = f(y)(f(\delta yf(y))-1) \implies \lim\limits_{\delta y \to 0} (f(y+\delta y)-f(y)) = \lim\limits_{\delta y \to 0} f(y)(f(\delta yf(y))-1) = f(y)(f(0)-1) =0 $
For that to be true you would have to have $f(x)$ be continuous in $0$. If $f(x)$ is discontinuous in $0$ it would be discontinuous everywhere, that could be.
But if it's continuous in $0$ it would have to be continuous everywhere. But your solution is not continuous in $x=2$