Polynomial Functional Equation: $f(x)f(y)+2=f(x)+f(y)+f(xy)$ 
Let $f(x)$ be a one-one, polynomial function such that  $f(x)f(y)+2=f(x)+f(y)+f(xy) , \ \forall x,y \in \mathbb R \setminus \{0\}$, $f(1) \ne 1$, $f'(1)=3$. Find $f(x)$.

I tried to find the degree of the polynomial from the equation by using suitable substitution, but it didn't work. Also, I found that $f(1)=2$ and then I substituted  $y=\frac{1}{x}$ to get $f(x)f\left(\frac{1}{x}\right)= f(x) + f\left(\frac{1}{x}\right)$. But I can't simplify further. Also, the answer given in my book is $f(x)=x^3+1$.
Any help will be appreciated.
Thanks.
 A: Here's something to get you started.
You know that $f(x)$ is a polynomial function, so it is also continuous. We also know that
$$f(x)f(y)+2=f(x)+f(y)+f(xy)$$
if both $x$ and $y$ are zero. However, taking both $x$ and $y$ to approach zero, the equation is also true for either $x$ or $y$ or both being zero.
Substitute $y=0$. Substituting and solving yields
$$(f(0)-1)f(x)=2f(0)-2$$
This could be true for all $x$ only if $f(0)=1$ or if $f(x)=\frac{2f(0)-2}{f(0)-1}$. This last would make $f(x)$ a constant polynomial, which is prohibited by $f'(1)=3$. Therefore, $f(0)=1$.
(After I finished my last edit for the last paragraph, I saw that @tampis had done basically the same using the previous version of my answer. He deserves due credit, but I did write this on my own!)
Similar substitution for $x=y=1$ gives you $f(1)=1$ or $f(1)=2$. The first is given to be false, so $f(1)=2$, as you already found.
A: To add the solution by @Rory Daulton: Set $y=0$ and you get $$f(x)f(0)+2=f(x)+2f(0)$$ With $f(0)=2$ you get $$f(x)=2f(0)-2$$ which contradicts $f^\prime(1)=3$. Thus $f(0)=1$.
EDIT 1: Take $\partial_x$ from both sides in $$f(x)f(y)+2=f(x)+f(y)+f(xy)$$ which yields $$f^\prime(x)f(y)=f^\prime(x)+yf^\prime(xy)$$
Set $x=1$:$$3f(y)=3+yf^\prime(y)$$ Thus $$3f(y)-3=yf^\prime(y)$$ From This ODE you can derive, that the degree of the polynomial is 3. Now you can solve this ODE...
