7. Given real numbers $x, y$ satisfy:
$$
x>\max \{-3, y\},(x+3)\left(x^{2}-y^{2}\right)=8 \text {. }
$$

Then the minimum value of $3 x+y$ is $\qquad$ .
7. $4 \sqrt{6}-6$.

From the problem, we know $x+3>0, x-y>0$.
Then $x+y=\frac{8}{(x+3)(x-y)}>0$.
Thus, $8=(x+3)(x+y)(x-y)$
$$
\begin{array}{l}
=\lambda(x+3) \mu(x+y)(x-y) \\
\leqslant\left(\frac{\lambda(x+3)+\mu(x+y)+(x-y)}{3}\right)^{3},
\end{array}
$$

where, $\lambda \mu=1, \lambda-2 \mu+4=0$.
Therefore, $\lambda=\sqrt{6}-2, \mu=\frac{\sqrt{6}+2}{2}$.
Substituting into equation (1) gives
$$
\begin{array}{l}
\left(\frac{\frac{\sqrt{6}}{2}(3 x+y)+3 \sqrt{6}-6}{3}\right)^{3} \geqslant 8 \\
\Leftrightarrow 3 x+y \geqslant 4 \sqrt{6}-6,
\end{array}
$$

and when $x=\sqrt{6}-1, y=\sqrt{6}-3$, the equality holds.