7. The natural domain of the function $y=f\left(\frac{2 x}{3 x^{2}+1}\right)$ is $\left[\frac{1}{4}, a\right]$, then the value of $a$ must be
7. $\frac{4}{3}$

Let $g(x)=\frac{2 x}{3 x^{2}+1}$, then the natural domain of $g(x)$ is $R$, and $g\left(\frac{1}{3 x}\right)=\frac{2 \cdot \frac{1}{3 x}}{3 \cdot \frac{1}{9 x^{2}}+1}=$
$$
\frac{2 x}{1+3 x^{2}}=g(x)(x \neq 0)
$$

Given $\frac{1}{4} \in\left[\frac{1}{4}, a\right]$, we know $-\frac{1}{3 \times \frac{1}{4}}=\frac{4}{3} \in\left[\frac{1}{4}, a\right]$, so $a \geqslant \frac{4}{3}$
However, $a>\frac{4}{3}$ does not meet the problem's requirements, because if there exists $x_{0}>\frac{4}{3}$ such that $f\left[g\left(x_{0}\right)\right]$ is defined, then $\frac{1}{3 x_{0}}$ $\left(\in\left(0, \frac{1}{4}\right)\right)$ would also make $f\left[g\left(\frac{1}{3 x_{0}}\right)\right]$ defined, but numbers less than $\frac{1}{4}$ cannot make $f(x)$ defined. Therefore, $a=\frac{4}{3}$.