5. The function $f(x)$ is defined on $(0, \infty)$, and satisfies
$$
f(x)-2 x f\left(\frac{1}{x}\right)+3 x^{2}=0 \text {. }
$$

Then the minimum value of $f(x)$ is $\qquad$
5.3.

From $f(x)-2 x f\left(\frac{1}{x}\right)+3 x^{2}=0$, we get $f\left(\frac{1}{x}\right)-\frac{2}{x} f(x)+\frac{3}{x^{2}}=0$.
By solving the above two equations simultaneously, we obtain $f(x)=x^{2}+\frac{2}{x}$.
By the AM-GM inequality,
$$
x^{2}+\frac{2}{x}=x^{2}+\frac{1}{x}+\frac{1}{x} \geqslant 3\left(x^{2} \cdot \frac{1}{x} \cdot \frac{1}{x}\right)^{\frac{1}{3}}=3 \text {, }
$$

and when $x=1$, the equality holds.
Therefore, the minimum value of $f(x)$ is 3.