I am searching for an unusual real-valued function.

It needs to fulfill following condition for all positive $x,y ∈ ℝ$: $$f(x)+f(y)+2xyf(xy)=\frac{f(xy)}{f(x+y)}$$

any ideas?

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Based on the answers below, I guess the real question is: does anything else work? – Semiclassical Jul 21 '14 at 1:51

I am sorry. Try as I might, I can't rule out all possibility. However, I managed to obtain a partial result: if we further assume that the function is continuous, then there is only 1 solution. This is because I can prove that there is only 1 possible solution for rational number. As for non-continuous solution, I have a good idea on how to extend uniqueness to Euclidean field, a vague hint on how to extend uniqueness to algebraic field, but completely no clues on transcendental numbers. In fact, I suspect that there might not be unique solution once you extend the issue to transcendental number, but I have not been able to find an example of such.

Set $x=y=1$ give us $4f(1)=\frac{f(1)}{f(2)}$ so $f(2)=\frac{1}{4}$.

Set $x=y=2$ give us $2f(2)+8f(4)=\frac{f(4)}{f(4)}=1$ show that $f(4)=\frac{1}{16}$.

Set $x=1,y=2$ give us $f(1)+5f(2)=\frac{f(2)}{f(3)}$ hence $f(1)+\frac{5}{4}=\frac{1}{4f(3)}$. While set $x=1,y=3$ give us $f(1)+7f(3)=\frac{f(3)}{f(4)}$ hence $f(1)+7f(3)=16f(3)$ hence $f(1)=9f(3)$. Plug back give us $9f(3)+\frac{5}{4}=\frac{1}{4f(3)}$ which give $36(f(3))^{2}+5f(3)-1=0$ so $f(3)=\frac{-5\pm 13}{72}=\frac{1}{9},-\frac{1}{4}$. The corresponding value for $f(1)$ is $1,-\frac{9}{4}$.

Set $x=1,y=n$ for any positive integer $n$ give $f(1)+(2n+1)f(n)=\frac{f(n)}{f(n+1)}$ so $f(n+1)=\frac{f(n)}{f(1)+(2n+1)f(n)}$, which for fixed $f(1)$, determine the sequence $f(n)$.

Consider case $f(1)=-\frac{9}{4}$. We continue the sequence to find $f(5)=-\frac{1}{27},f(6)=\frac{4}{287}$

Set $x=2,y=3$ give us $f(2)+f(3)+12f(6)=\frac{f(6)}{f(5)}$. Plug in the above value give $\frac{1}{4}-\frac{1}{4}+12\frac{4}{287}=\frac{(\frac{4}{287})}{(-\frac{1}{27})}$ hence $\frac{48}{287}=-\frac{108}{287}$ contradiction. Hence $f(1)=1$ and $f(3)=\frac{1}{9}$.

Set $x=1$ and let $y$ be arbitrary we get $f(1)+f(y)+2yf(y)=\frac{f(y)}{f(n+y)}$ so $1+(2y+1)f(y)=\frac{f(y)}{f(y+1)}$ so $f(y+1)=\frac{f(y)}{1+(2y+1)f(y)}$.

Now a simple result make things easier: $\frac{(\frac{z}{1+az})}{1+b(\frac{z}{1+az})}=\frac{z}{1+(a+b)z}$. This leads by induction to $f(y+n)=\frac{f(y)}{1+f(y)\sum\limits_{k=0}^{n-1}(2(y+k)+1)}=\frac{f(y)}{1+(2y+n)nf(y)}$ (this formula will be used repeatedly, so remember this). In particular, plug in $y=1$ give us the formula $f(n)=f(1+(n-1))=\frac{1}{1+(2+n-1)(n-1)}=\frac{1}{n^{2}}$.

Take $x=n,y=\frac{1}{n}$ we get $\frac{1}{n^{2}}+f(\frac{1}{n})+2=\frac{1}{f(n+\frac{1}{n})}=\frac{1+(\frac{2}{n}+n)nf(\frac{1}{n})}{f(\frac{1}{n})}=\frac{1}{f(\frac{1}{n})}+2+n^{2}$ so $f(\frac{1}{n})-\frac{1}{f(\frac{1}{n})}=n^{2}-\frac{1}{n^{2}}$ which have only solution $f(\frac{1}{n})=n^{2}$ or $f(\frac{1}{n})=-\frac{1}{n^{2}}$

Take $x=2n,y=\frac{1}{n}$ we get $\frac{1}{4n^{2}}+f(\frac{1}{n})+1=\frac{1}{4f(2n+\frac{1}{n})}=\frac{1}{4}\frac{1+(\frac{2}{n}+2n)2nf(\frac{1}{n})}{f(\frac{1}{n})}=\frac{1}{4f(\frac{1}{n})}+1+n^{2}$ so $f(\frac{1}{n})-\frac{1}{4f(\frac{1}{n})}=n^{2}-\frac{1}{4n^{2}}$ which once again have only solution $f(\frac{1}{n})=n^{2}$ or $f(\frac{1}{n})=-\frac{1}{4n^{2}}$

Combine the above 2 results we get $f(\frac{1}{n})=n^{2}$. In other word, when $x=\frac{1}{n}$ we have $f(x)=\frac{1}{x^{2}}$.

Apply the formula with $y=\frac{1}{k}$ we get $f(n+\frac{1}{k})=\frac{f(\frac{1}{k})}{1+(\frac{2}{k}+n)nf(\frac{1}{k})}=\frac{k^{2}}{1+(\frac{2}{k}+n)nk^{2}}=\frac{k^{2}}{1+2nk+(nk)^{2}}=\frac{k^{2}}{(nk+1)^{2}}=\frac{1}{(n+\frac{1}{k})^{2}}$. Thus for $x$ of the form $n+\frac{1}{k}$ we get $f(x)=\frac{1}{x^{2}}$.

Now to find arbitrary $f(\frac{p}{q})$. Set $x=p,y=\frac{1}{q}$. We get $f(p)+f(\frac{1}{q})+2\frac{p}{q}f(\frac{p}{q})=\frac{f(\frac{p}{q})}{f(p+\frac{1}{q})}$ hence $\frac{1}{p^{2}}+q^{2}+2\frac{p}{q}f(\frac{p}{q})=(p+\frac{1}{q})^{2}f(\frac{p}{q})$ so $\frac{1}{p^{2}}+q^{2}=(p^{2}+\frac{1}{q^{2}})f(\frac{p}{q})$ so $f(\frac{p}{q})=\frac{\frac{1}{p^{2}}+q^{2}}{p^{2}+\frac{1}{q^{2}}}=(\frac{q}{p})^{2}$.

Hence for $x$ rational then $f(x)=\frac{1}{x^{2}}$.

Now if we assume that the function is continuous the only possibility is that $f(x)=\frac{1}{x^{2}}$ everywhere. Then for any $x,y$ we have $f(x)+f(y)+2xyf(xy)=\frac{1}{x^{2}}+\frac{1}{y^{2}}+2xy\frac{1}{x^{2}y^{2}}=\frac{x^{2}+y^{2}+2xy}{x^{2}y^{2}}=\frac{(x+y)^{2}}{(xy)^{2}}=\frac{f(xy)}{f(x+y)}$. Thus this is indeed a solution.

Hence the only continuous solution is $f(x)=\frac{1}{x^{2}}$.

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$f(x)=\frac{1}{x^2}$ is a solution.

I came up with this by setting $x=y=1$ and finding $f(2)=\frac{1}{4}$.

Towards a proof of uniqueness:

It is trivial to show that

$$f(x+1) = \frac{1}{1+2x+\frac{1}{f(x)}}$$

from which it follows that

\begin{align} f(x)&>\frac{1}{x^2} \Rightarrow f(x+1)>\frac{1}{(x+1)^2} \\ f(x)&<\frac{1}{x^2} \Rightarrow f(x+1)<\frac{1}{(x+1)^2} \end{align}

I'm so close I can taste it.

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wow - that's astonishing! thanks – user161516 Jul 21 '14 at 1:31
This answer is undefined for $x=0$, contradict the requirement that the equality work for all $x,y\in\mathbb{R}$. – Gina Jul 21 '14 at 1:56
@Gina "for all positive [..]" – user161516 Jul 21 '14 at 2:00
@Levix: oh right sorry. – Gina Jul 21 '14 at 2:02
When it come to proving uniqueness, the main difficulty seems to be transcendental number: sure you might have the above equality, but 1 equality is easily satisfied, and there does not seems to be anyway to obtain any more to cause further restriction, as the number is transcendental. – Gina Jul 21 '14 at 14:49

Set $h(x) = 1/f(x)$ and assume $f(xy) = f(x)f(y)$, then we see that

$$h(x) + h(y) + 2xy = h(x+y)$$

Thus $h(x) = x^2$ is a solution.

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oops, ozo posted this also. I guess posting another approach should be ok. – Knoothe Jul 21 '14 at 1:43
-1 I don't think you can "assume $f(xy) = f(x) f(y)$", but otherwise this is nice. – Calvin Lin Jul 21 '14 at 3:34
@CalvinLin It's a guess that produces a valid solution. – Casey Chu Jul 21 '14 at 4:05