Let
$$
f(x)=a_0 x^n+a_1 x^{n-1}+a_2 x^{n-2}+\ldots+a_{n-1} x+a_n
$$

Then, $f(x) \cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$
$$
\begin{aligned}
\Rightarrow\left(a_0 x^n+a_1 x^{n-1}+\ldots+a_n\right) & \left(\frac{a_0}{x^n}+\frac{a_1}{x^{n-1}}+\ldots+a_n\right)
\end{aligned}
$$

On comparing the coefficient of $x^n$, we have
$$
a_0 a_n=a_0 \Rightarrow a_n=1
$$

Comparing the coefficient of $x^{n-1}$, we have
$$
\begin{aligned}
& \Rightarrow a_0 a_{n-1}+a_n a_1=a_1 \\
& \Rightarrow \quad a_0 a_{n-1}+a_1=a_1 \\
& {\left[\text { as } a_n=1\right]} \\
& \Rightarrow \quad a_0 a_{n-1}=0 \\
& \Rightarrow \quad a_{n-1}=0 \quad\left[\text { as } a_0 \neq 0\right] \\
&
\end{aligned}
$$

Similarly, $\quad a_{n-1}=a_{n-2}=\ldots=a_1=0$
and $\quad a_0= \pm 1$
$$
\begin{aligned}
& \therefore \quad f(x)=1 \pm x^n \\
& f(4)=1 \pm 4^n=257 \Rightarrow 4^n=256 \\
& \Rightarrow \quad 4^n=256 \quad \Rightarrow \quad 4=4 \\
& f(x)=1+x^4 \\
&
\end{aligned}
$$

So, $f(3)=1+3^4=82$