6. Let $f_{1}(x)=-\frac{2 x+7}{x+3}, f_{n+1}(x)=f_{1}\left(f_{n}(x)\right), x \neq-2, x \neq-3$, then $f_{2022}(2021)=$
$$
f_{2}(x)=-\frac{2 \cdot\left(-\frac{2 x+7}{x+3}\right)+7}{-\frac{2 x+7}{x+3}+3}=-\frac{3 x+7}{x+2}, f_{3}(x)=-\frac{2 \cdot\left(-\frac{3 x+7}{x+2}\right)+7}{-\frac{3 x+7}{x+2}+3}=x \text {, }
$$

Therefore, $f_{2022}(x)=f_{3 \times 674}(x)=x \Rightarrow f_{2022}(2021)=2021$.