Let $y=\frac{x+1}{x}$, then $x=\frac{1}{y-1}$. It follows $$f\Big(\frac{x+1}{x}\Big)= 1 + \frac{1}{x} + \frac{1}{x^2}$$ $$\implies f(y)=1+(y-1)+(y-1)^2 = y^2 - y +1$$ Therefore the answer is $$f(x)=\boxed{x^2-x+1}$$