Finding the value of $f(1)$ from a given functional equation 
A function $f: \mathbb{Q}^+ \cup \{0\} \to \mathbb{Q}^+ \cup \{0\}$ is defined such that $$ f(x) + f(y) + 2xyf(xy) = \frac{f(xy)}{f(x+y)}$$
  Then what is the value of $\left[f(1)\right]$ (where $[.]$ denotes the greatest integer function)?  

I proceeded this way:
Putting $x=y=0$ I got $f(0) = \frac{1}{2}$ (assuming $f(0) \neq 0$)
Again putting $y=0$ I got $f(x) + f(0) = \frac{f(0}{f(x)}$ which gave 2 values of $f(x)$ as $-1$ and $\frac{1}{2}$.
As $f(0)$ was equal to $\frac{1}{2}$ so I assumed $f(x)$ as a constant function having value $\frac{1}{2}$ for all $x$.
When $f(0) = 0$ then I got $f(x) = 0$ for all $x$.
But the answer was given to be equal to $1$ which means $f(1) \in [1,2)$. Where am I wrong?
 A: Your logic is fine.  You can rule out $f(0)=0$ because of the divide by zero it produces on the right.  You can rule out $f(x)=-1$ because range of the function is given as $\Bbb Q \cup \{0\}$.  Then you can say for $x=y=1$ the defining equation asserts $\frac 12+\frac 12+2\cdot 1 \cdot 1 \cdot \frac12=\dfrac{(\frac 12)}{(\frac 12)}$, which is false.  There is no such function.
A: *

*Assume $f(0) \neq 0$.
Then $f(0) + f(0) = \frac{f(0)}{f(0)} \rightarrow f(0) = \frac{1}{2}$,
and $f(x) + f(0) = \frac{f(0)}{f(x)} \rightarrow f(x)^2 + \frac{1}{2}f(x) -\frac{1}{2} = 0.$


*Assume $f(1) \neq 0$.
We also know that $f(1) + f(1) + 2f(1) = \frac{f(1)}{f(2)}$. Or $f(2) = \frac{1}{4}$.
However, $(\frac{1}{4})^2 + \frac{1}{2}\cdot\frac{1}{4} - \frac{1}{2} \neq 0$. Contradiction.

*Assume $f(1) = 0$.
$0^2 + \frac{1}{2}\cdot 0 - \frac{1}{2} \neq 0$. Contradiction.
Therefore we have $f(0) = 0$. However then we have for any $x$:
$$f(x) = \frac{0}{f(x)}$$
If $f(x) \neq 0$ then the above formula implies $f(x)^2 = 0$ which is a contradiction.
So the function must be $f(x) = 0$ everywhere however if this is the case then the 'definition' is ill-defined (division by zero) everywhere, therefore I argue that $f$ itself is ill-defined.
