Given f(x)=65 and f(x)f(1/x)=f(x)+f(1/x)... find the value of f(6)=???
There is not enough information here for the problem to have a unique solution. I suspect that the intended answer is that $\displaystyle f(x) = x^3+1$, and so f(6) = 217. But in fact if g(x) is an arbitrary function such that g(1/x) = g(x) and g(4)=3, then $\displaystyle f(x) = x^{g(x)}+1$ is a solution, and f(6) can be a completely arbitrary number greater than 1.
The equation $\displaystyle f(x)f(1/x)=f(x)+f(1/x)$ can be written as $\displaystyle \bigl((f(x)-1\bigr)\bigl((f(1/x)-1\bigr) = 1.$ So if $\displaystyle g(x) = f(x)-1$ then $\displaystyle g(1/x) = 1/g(x)$. There are very many functions having that property, but the most obvious solutions are $\displaystyle g(x) = x^k$ for some fixed k.
If $\displaystyle f(4) = 65$ then $\displaystyle g(4) = 64 = 4^3$. That suggests that we should take k=3, giving $\displaystyle g(x) = x^3$ and hence $\displaystyle f(x) = x^3+1.$
But as I said in the previous comment, there is nothing unique about that solution. If there really was no further information given, then this is a very badly thought out problem.