Solution:

The given equations are
$\begin{array}{l}
\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\dots \dots(i) \\
\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8}
\end{array}$
$\frac{1}{3 x+y}-\frac{1}{3 x-y}=-\frac{1}{4} \text { (Multiplying by 2) } \ldots \ldots \text { (ii) }$
Substituting $\frac{1}{3 x+y}=u$ and $\frac{1}{3 x-y}=v$ in equation(i) and (ii), we obtain:
$\begin{array}{l}
\mathrm{u}+\mathrm{v}=\frac{3}{4}\dots \dots(iii) \\
\mathrm{u}-\mathrm{v}=-\frac{1}{4}\dots \dots(iv)
\end{array}$
Adding equation(iii) and equation(iv), we obtain:
$2 \mathrm{u}=\frac{1}{2}$
$\begin{array}{l}
\Rightarrow \mathrm{u}=\frac{1}{4} \\
\Rightarrow 3 \mathrm{x}+\mathrm{y}=4 \quad\left(\because \frac{1}{3 x+y}=u\right)\dots \dots(v)
\end{array}$
Now, substituting $\mathrm{u}=\frac{1}{4}$ in equation(iii), we obtain:
$\begin{array}{l}
\frac{1}{4}+\mathrm{v}=\frac{3}{4} \\
\mathrm{v}=\frac{3}{4}-\frac{1}{4} \\
\Rightarrow \mathrm{v}=\frac{1}{2} \\
\Rightarrow 3 \mathrm{x}-\mathrm{y}=2 \quad\left(\because \frac{1}{3 x-y}=v\right) \quad \ldots .(\mathrm{vi})
\end{array}$
Adding equation(v) and equation(vi), we obtain
$6 x=6 \Rightarrow x=1$
Substituting $\mathrm{x}=1$ in equation(v), we have
$3+y=4 \Rightarrow y=1$
As a result, $x=1$ and $y=1$.