Find $f(x)$ such that $f(x)+f\left(\frac{1}{x}\right)=f(x)\cdot f\left(\frac{1}{x}\right)$

Hint: $$f(x) = 1 + \frac{1}{f(\frac{1}{x})-1}$$

So $f(x) - 1 = \frac{x^n}{P(x)}$, where P(x) is a polynomial of degree $\le n$. The left-hand side is a polynomial, so $P(x) = a x^k$ for some $k$. Now $a$ and $k$ can be found from the functional equation itself. But, as Barry Cipra notes in comments, since the left-hand side of the equality that defines $P$ has degree $n$ the only possible value for $k$ is $0$.

Assume $f(x)=c$ is constant. Then we arrive at $c+c=c\cdot c$, i.e. $c=0$ or $c=2$.

Otherwise write $f(x)=c+x^kg(x)$ where $k\ge1$ and $g(0)\ne0$. Then for $x\ne0$ $$2c+x^kg(x)+x^{-k}g(x^{-1})=f(x)+f(\frac1x)=f(x)f(\frac1x)=cx^kg(x)+c^2+cx^{-k}g(x^{-1})+g(x)g(\frac1x)$$ From looking at the highest power occuring in $g$ we see $c=1$, hence this simplifies to $$ g(x)g(\frac1x)=1$$ so $g$ is a monomial, hence constant. Thus $f(x)=1+ax^k$ and we further check that $a=\pm1$. Thus the complete list of solutions is $$ f(x)=0,\qquad f(x)=2,\qquad f(x)=1\pm x^k\text{ with }k\ge1$$

$$f(y) = 1+y$$

$$f(x)+f\left(\frac{1}{x}\right) = 1+x+1+\frac{1}{x} = 2 + x + \frac{1}{x}$$

$$f(x) f\left(\frac{1}{x}\right) = (1+x) \left ( 1+\frac{1}{x}\right) = 1 + x + \frac{1}{x} + x \frac{1}{x} = 2 + x + \frac{1}{x}$$

In fact, $f(y) = 1+y^k$ for $k \in \mathbb{N}$ seems to work for the same reason.