Solution:
The domain of $f(x)$ is $x > 0$. Setting $y = \lg x$ gives
$$F(y) = \frac{2y^{2} + 3y + 3}{y^{2} + 2y + 2}$$
Since the denominator is positive, the function $F(y)$ is defined for all real $y$.
Let $M$ be the desired value of $f(x)$ (if it exists). Then for any real $y$ we have
$$\begin{gathered} \frac{2y^{2} + 3y + 3}{y^{2} + 2y + 2} \leq M \\ 2y^{2} + 3y + 3 \leq M y^{2} + 2M y + 2M \\ (2 - M) y^{2} + (3 - 2M) y + (3 - 2M) \leq 0 \end{gathered}$$
Therefore $2 - M < 0$ and $D = (3 - 2M)(3 - 2M - 8 + 4M) = (3 - 2M)(2M - 5) \leq 0$. Hence $M \geq 2.5$.
Note that for $M = 2.5$ the above inequality becomes $-0.5 y^{2} - 2y - 2 \leq 0$, i.e. $y^{2} + 4y + 4 = (y + 2)^{2} \geq 0$ and the equality is attained only if $y = -2$, i.e. for $x = 0.01$.
Therefore the maximum of the function equals $2.5$ and it is attained for $x = 0.01$ only.