If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$ Problem : 
If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$ 
Solution : 
$f(x) f(\frac{1}{x})-f(x) =f(\frac{1}{x})$
$\Rightarrow f(x) =\frac{f(1/x)}{f(1/x)-1}$.....(i)
Also $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$
$\Rightarrow f(\frac{1}{x})=\frac{f(x)}{f(x)-1}$ ......(ii)
On multiplying (i) and (ii) , we get 
$f(x) .f(\frac{1}{x})=\frac{f(1/x).f(x)}{(f(1/x)-1) ((f)(x)-1)}$
$\Rightarrow (f(\frac{1}{x}) -1)(f(x)-1)=1$ 
Please suggest how to proceed here since $f(x)-1 ; \& f(\frac{1}{x}-1)$ are reciprocal to each other 
Thanks

Let $f(x)=\sum_{k=0}^na_kx^k,f\neq0$, where $n$ is the degree of $f$, i.e. $a_n\neq0$. If $n=0$, then your equation is equivalent to $a_0^2=2a_0$, which implies $a_0=2$. So let's assume that $n\geq1$. Multiply both sides of your equation by $x^n$ and use the Cauchy product formula for finite sums. This yields after some simplifications
\begin{align*}
\sum_{j=0}^{n-k}a_ja_{j+k}&=a_k,\qquad\forall k\in\{1,\ldots,n\}, \\
\sum_{j=0}^na_j^2&=2a_0.
\end{align*}
For $k=n$ we get $a_0a_n=a_n$, and since $a_n\neq0$ we have $a_0=1$.
For $k=n-1$ we get $a_0a_{n-1}+a_1a_n=a_{n-1}$, which is equivalent to $a_1a_n=0$ and hence $a_1=0$. For $k=n-2$ we obtain $a_0a_{n-2}+a_1a_{n-1}+a_2a_n=a_{n-2}$, thus $a_2a_n=0$ and finally $a_2=0$. Continuing this process gives $a_k=0$ for each $k\in\{1,\ldots,n-1\}$. From the last equality we then have that $1+a_n^2=2$, which implies $a_n=\pm1$. Consequently, $f$ is of the form $f(x)=1\pm x^n$.

Using $f(4)=65$, we conclude $f(x)=x^3+1$ (see George V. Williams' answer).
