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I was studying the functional equation $f(xf(y))f(y)=f(x+y)$ and observed that for $x=\delta y$ we can write $f(y+\delta y)-f(y) = f(y)(f(\delta yf(y))-1)$. 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 $f(x)= \frac{2}{2-x}$ $x\ne2$

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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 $x=2$

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    Thanks, incidentally the original problem actually includes a few more conditions restricting the domain to real numbers < 2, so it all works out despite 2 not being in the domain of 2/(x-2). So is there no other way to prove continuity of the function? – 2017-02-17
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    @CompulsiveMathurbator Have you found your solution without assuming continuity? In that case the proof would be in there somewhere.. – 2017-02-17
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    as a matter of fact yes, the same inequalities I used to solve it can prove continuity (redundant but possible). Thanks. – 2017-02-17