f(x)+f(1/x)=f(x)f(1/x)[f(x)-1][f(1/x)-1]=1

Let f(x)-1=g(x) then g(x).g(1/x)=1.

If g(x) is a constant function k then k²=1 which gives f(x)=0 or f(x)=2.

Let g(x)=a₀+a₁x+a₂x²+...+a_nxⁿ ,a nth degree polynomial(a_n0) then

g(1/x)=a₀+a₁/x+a₂/x²+...+a_n/xⁿ

g(x).g(1/x)=1coeff. of xⁿ=a₀a_n=0a₀=0.

So,g(x)=x[a₁+....+a_nx^n-1].Again g(x).g(1/x)=1 gives a₁a_n=0a₁=0

If we proceed like this,we get,a₀=a₁=a₂=...=a_n-1=0.

So,g(x)=a_nxⁿ where a_n²=1 or a_n=1.

Hence,f(x) can be the following

f(x)=0 or f(x)=2 or f(x)=1xⁿ.