(2) When $0<x<1$, $f(x)=\frac{x}{\lg x}$, then the correct size relationship is ( ).
(A) $f^{2}(x)<f\left(x^{2}\right)<f(x)$
(B) $f\left(x^{2}\right)<f^{2}(x)<f(x)$
(C) $f(x)<f\left(x^{2}\right)<f^{2}(x)$
(D) $f\left(x^{2}\right)<f(x)<f^{2}(x)$
(2) When $00$.
Also, since
$$
\frac{x}{\lg x}-\frac{x^{2}}{\lg x^{2}}=\frac{2 x-x^{2}}{2 \lg x}=\frac{(2-x) x}{2 \lg x}<0 .
$$

Therefore,
$$
f(x)<f\left(x^{2}\right)<f^{2}(x) .
$$

So the answer is C.