Question -
Find all $f : \mathbb{R} \to\mathbb{R}$ such that
$f(x+y)=f(x)f(y)f(xy)$ for all $x,y \in \mathbb{R}$.
My try -
Putting $x=y=0$ I get three cases $f(0)=0$,$f(0)=1$,$f(0)=-1$
For $f(0)=0$ we get by putting $y=0$ in equation, $f(x)=0$ for all $x$ ...I guess other two solutions are $f(x)=1$ and $-1$ but not able to prove it using induction on there corresponding values of $f(0)$...
Any hints ?
Suppose there exists a number $a$ such that $f(a)=0$. Plugging $P(x-a, a)$ in the equation we get $f\equiv 0$, whichis our first solution. Otherwise, by comparing $P\left(x, y+\frac{1}{xy}\right)$ with $P\left(x+\frac{1}{xy}, y\right)$ and using the given equation, we get $$f\left(x+y+\frac{1}{xy}\right)=f(x)f\left(y+\frac{1}{xy}\right)f\left(xy+\frac{1}{y}\right)=f(x)f(y)f\left(\frac{1}{xy}\right)f\left(\frac{1}{x}\right)f(xy)f\left(\frac{1}{y}\right)f(x)$$ and $$f\left(x+y+\frac{1}{xy}\right)=f(x)f(y)f\left(\frac{1}{xy}\right)f\left(\frac{1}{x}\right)f(xy)f\left(\frac{1}{y}\right)f(y),$$ which gives $$f(x)=f(y)\ \forall\ x,y\ne 0,$$ so $f(x)=c$ for nonzero $x$. Plugging $P(1, -1)$ in the equation we get $f(0)=c^3=c$ and so the only solutions are $$\boxed{f\equiv 0}\ \boxed{f\equiv 1}\ \boxed{f\equiv -1}$$
Some partial result.
If $y=0$ we get $$f(x)= f(x)f(0)^2 \implies f(0)=\pm1\;\;\vee \;\;f(x)=0,\;\forall x$$
So one solution is $f(x)=0$ for all $x$.
Now, say exists $c$ such that $f(c)\ne 0$.
Then if $f$ has at least one zero $r$ (which can not be $0$) then we have $$f(x+r) = f(x)f(r)f(xr)=0$$ so again we have $f(x)=0$ for all $x$. A contradiction. So $f$ has no zeros.
Now let $y={x\over x-1}$ then we have $f(x+y)= f(xy)$ so we get $$1 = f(x)f({x\over x-1})$$ for all $x\ne 1$ so $$\boxed{f\Big({x\over x-1}\Big) = {1\over f(x)};\;\;\forall x\ne 1}$$
Also, for $y=-x$ we get $$f(0) = f(x)f(-x)f(-x^2)$$
