6. Given $0 \leqslant 6 x 、 3 y 、 2 z \leqslant 8$,
$$
\sqrt{12 x}+\sqrt{6 y}+\sqrt{4 z}=6 \text {. }
$$

Then the function
$$
f(x, y, z)=\frac{1}{1+x^{2}}+\frac{4}{4+y^{2}}+\frac{9}{9+z^{2}}
$$

has a maximum value of $\qquad$ .
6. $\frac{27}{10}$.

Let $y=2 y_{1}, z=3 z_{1}$, then
$$
0 < x, y_{1}, z_{1} \leqslant \frac{4}{3}, \sqrt{\frac{x}{3}}+\sqrt{\frac{y_{1}}{3}}+\sqrt{\frac{z_{1}}{3}}=1,
$$

and $f(x, y, z)=\frac{1}{1+x^{2}}+\frac{1}{1+y_{1}^{2}}+\frac{1}{1+z_{1}^{2}}$.
By $1=\sqrt{\frac{x}{3}}+\sqrt{\frac{y_{1}}{3}}+\sqrt{\frac{z_{1}}{3}}$
$$
\leqslant \frac{1}{2}\left(x+\frac{1}{3}\right)+\frac{1}{2}\left(y_{1}+\frac{1}{3}\right)+\frac{1}{2}\left(z_{1}+\frac{1}{3}\right),
$$

we get $x+y_{1}+z_{1} \geqslant 1$.
By symmetry, we guess: when $x=y_{1}=z_{1}=\frac{1}{3}$, $f(x, y, z)$ reaches its maximum value $\frac{27}{10}$.

Let $\frac{1}{1+x^{2}} \leqslant A x+B$, where, when $x=\frac{1}{3}$, the equality holds.
$$
\begin{array}{l}
\text { By }(A x+B)\left(1+x^{2}\right)-1 \geqslant 0 \text {, and } \\
(3 x-1)^{2}(a x+b) \geqslant 0,
\end{array}
$$

comparing coefficients, we get $A=-\frac{27}{50}, B=\frac{27}{25}$. Thus,
$$
\begin{array}{l}
\frac{1}{1+x^{2}} \leqslant \frac{27}{25}\left(1-\frac{1}{2} x\right) \\
\Leftrightarrow(3 x-1)^{2}\left(x-\frac{4}{3}\right) \leqslant 0 .
\end{array}
$$

Similarly, $\frac{1}{1+y_{1}^{2}} \leqslant \frac{27}{25}\left(1-\frac{1}{2} y_{1}\right)$,
$$
\frac{1}{1+z_{1}^{2}} \leqslant \frac{27}{25}\left(1-\frac{1}{2} z_{1}\right).
$$

Therefore, $f(x, y, z) \leqslant \frac{27}{25}\left[3-\frac{1}{2}\left(x+y_{1}+z_{1}\right)\right]$
$$
\leqslant \frac{27}{25}\left(3-\frac{1}{2} \times 1\right)=\frac{27}{10},
$$

where, when $x=y_{1}=z_{1}=\frac{1}{3}$, the equality holds.
Thus, the maximum value of $f(x, y, z)$ is $\frac{27}{10}$.