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 $f(36)=39$.
Notice that $$\frac{f(0)f(0)-f(0^2)}{3}=0+0+2\Rightarrow f^2(0)-f(0)=6\Rightarrow f(0)=3 \hspace{0.5cm}\text{or}\hspace{0.5cm}f(0)=-2$$ Now set $y=0$ and $x\in\mathbb{R}$. You have $$\frac{f(x)f(0)-f(0)}{3}=x+2\Rightarrow f(x)=\frac{1}{f(0)}(3x+6+f(0))$$ Substitute for $f(0)$ the two possible cases and you are done. Then you can easily find the possible values of $f(36)$.
