$P(x,x)$ implies $f(0)=0$. Since $f(x)-f(1)x$ also satisfies the f.e., we will consider $f(x)-f(1)x$ instead of $f(x)$. Now we have $f(1)=0$. Comparing $P(x,y)$ with $P(y,x)$ implies $f(-x)=-f(x)$. $P(x,-x)$ implies $f(2x)=2f(x)$. So $f(2)=0$. $P(x+1,x-1)+P(x+1,1-x)$ yields $f(x+1)=\frac{x^2+4x-1}{(x+1)^2}f(x)$ when $x\neq -1$. Multiply by $2$ and replace $x$ with $x/2$ above, we get for $x\neq-2$, $f(x+2)=\frac{x^2+8x-4}{(x+2)^2}f(x)$ On the other hand, we have for $x\neq-1,-2$, $f(x+2)=\frac{f(x+2)}{f(x+1)}f(x+1)=\frac{(x^2+4x-1)(x^2+6x+4)}{(x+1)^2(x+2)^2}f(x)$. The two equations above implies $f(x)=0,\forall x\neq -1,-2,-\frac57$. Actually for any $y$, we can plug in a large $x$ to $P(x,y)$ and prove $f(y)=0,\forall y$. Since we replace $f(x)$ with $f(x)-f(1)x$ in the beginning, the final solution is $f(x)=f(1)x$. BTW, the boundedness condition seems unnecessary.

thanks, hmm... as it appears boundedness is a redundant requirement.

With the bounded criteria one can prove that f is Continuous and also derivable and find it by a differential equation. i also think that with this criteria we can use the fact that f is additive and prove that f is linear mapping.