(i) $x=(-1)$

$\begin{array}{l}
\text { L.H.S. }=x^{2}+2 x-1 \\
=3 \times(-1)^{2}+2 \times(-1)-1 \\
=3-2-1 \\
=0 \\
=\text { R.H.S. }
\end{array}$

Thus, $(-1)$ is a root of $\left(3 x^{2}+2 x-1=0\right)$.

(ii) On subtracting $x=\frac{1}{3}$ in the given equation, we get:

$\begin{array}{l}
\text { L.H.S. }=3 x^{2}+2 x-1 \\
=3 \times\left(\frac{1}{3}\right)^{2}+2 \times \frac{1}{3}-1 \\
=3 \times \frac{1}{9}+\frac{2}{3}-1 \\
=\frac{1+2-3}{3} \\
=\frac{0}{3} \\
=0 \\
=\text { R.H.S. }
\end{array}$

Thus, $\left(\frac{1}{3}\right)$ is a root of $\left(3 x^{2}+2 x-1=0\right)$,