8. If $a^{2}+b^{2}=1(a, b \neq 0)$,
$$
y=a+b+\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b} \text {, }
$$

then the minimum value of $|y|$ is
8. $2 \sqrt{2}-1$.

Notice that $y=a+b+\frac{1}{a b}+\frac{a+b}{a b}$.
Let $t=a+b$. Then, $a b=\frac{1}{2}\left(t^{2}-1\right)$, where $|t| \leqslant \sqrt{2}$, and $t \neq 1$.
Thus, $y=t+\frac{2}{t^{2}-1}+\frac{2 t}{t^{2}-1}$
$$
=t+\frac{2}{t-1}=t-1+\frac{2}{t-1}+1 \text {. }
$$

When $t>1$, $t-1+\frac{2}{t-1} \geqslant 2 \sqrt{2}$, at this time,
$$
y \geqslant 2 \sqrt{2}+1 \text {; }
$$

When $t<1$, $-y=1-t+\frac{2}{1-t}-1 \geqslant 2 \sqrt{2}-1$.
Therefore, $|y| \geqslant 2 \sqrt{2}-1$, and the equality holds when $t=1-\sqrt{2}$.
Thus, the minimum value of $|y|$ is $2 \sqrt{2}-1$.