<s>[INST] 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 $f(36)=f(4)f(9)-45$.
$$\frac{f(6)f(6)-f(6\cdot6)}{3}=6+6+2$$
$$\frac{f(2)f(18)-f(2\cdot18)}{3}=2+18+2$$
$$\frac{f(3)f(12)-f(3\cdot12)}{3}=3+12+2$$
$$\frac{f(36)f(1)-f(36\cdot1)}{3}=36+1+2$$
 [/INST] 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 $f(36)$.
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