10. The function $f(x)$ satisfies: for any real numbers $x, y$, we have
$$
\frac{f(x) f(y)-f(x y)}{3}=x+y+2 .
$$

Then the value of $f(36)$ is
10.39.

Let $x=y=0$, then $\frac{f^{2}(0)-f(0)}{3}=2$.
Therefore, $f(0)=-2$ or $f(0)=3$.
If $f(0)=-2$, let $y=0$, then $\frac{f(x) f(0)-f(0)}{3}=x+2$. So, $f(x)=-\frac{3}{2} x-2$. Substituting into the original equation shows it does not satisfy the condition.

If $f(0)=3$, solving yields $f(x)=x+3$, substituting for verification shows it satisfies the condition.
Thus, $f(36)=39$.