Finding $f(36)$ given $\frac{f(x)f(y)-f(xy)}{3} = x+y+2$ on $\mathbb R$ 
Let $f:\mathbb R \to \mathbb R$ be a function with
$$\frac{f(x)f(y)-f(xy)}{3} = x+y+2$$
for all real numbers $x,y$. List all possible values for $f(36)$.

So far I have just been plugging in possible $x$ and $y$.
$$\frac{f(4)f(9)-f(4\cdot9)}{3}=4+9+2$$
So then $f(36)=f(4)f(9)-45$.
$$\frac{f(6)f(6)-f(6\cdot6)}{3}=6+6+2$$
$$\frac{f(2)f(18)-f(2\cdot18)}{3}=2+18+2$$
$$\frac{f(3)f(12)-f(3\cdot12)}{3}=3+12+2$$
$$\frac{f(36)f(1)-f(36\cdot1)}{3}=36+1+2$$

If you let $x=y=0$ you get $f(0)^2-f(0) = 6$, which has solutions $f(0)=-2$ or $f(0)=3$. In both cases, setting $x=0$ gives
$$f(0)f(y)-f(0) = 3y+6 $$
$$f(y) = \frac{3y+6+f(0)}{f(0)}$$
Inserting $y=36$ and the two possible values of $f(0)$ then gives $f(36)=-56$ or $f(36)=39$.
