If $f\left(x+\dfrac{1}{x}\right)=x^2+\dfrac{1}{x^2}$ then $f(x)=?$

Let $x+\dfrac{1}{x}=z$.

Then,

$f(z)=f\left(x+\dfrac{1}{x}\right)=\left(x^2+\dfrac{1}{x^2}\right)=\left(x+\dfrac{1}{x}\right)^2-2=(z^2-2)$.

$\Rightarrow f(x)=(x^2-2)$.