Solve the following pair of equations by reducing them to a pair of linear equations:
$\frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4}$
$\frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = -\frac{1}{8}$

(N/A) Let $u = \frac{1}{3x+y}$ and $v = \frac{1}{3x-y}$.
Substituting these into the given equations:
$u + v = \frac{3}{4}$ --- $(1)$
$\frac{1}{2}u - \frac{1}{2}v = -\frac{1}{8}$ --- $(2)$
Multiply equation $(2)$ by $2$ to simplify:
$u - v = -\frac{1}{4}$ --- $(3)$
Adding equation $(1)$ and $(3)$:
$(u + v) + (u - v) = \frac{3}{4} - \frac{1}{4}$
$2u = \frac{2}{4} = \frac{1}{2}$
$u = \frac{1}{4}$
Substitute $u = \frac{1}{4}$ into equation $(1)$:
$\frac{1}{4} + v = \frac{3}{4}$
$v = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$
Now,solve for $x$ and $y$:
$3x+y = \frac{1}{u} = 4$ --- $(4)$
$3x-y = \frac{1}{v} = 2$ --- $(5)$
Adding $(4)$ and $(5)$:
$6x = 6 \implies x = 1$
Substitute $x = 1$ into $(4)$:
$3(1) + y = 4 \implies y = 1$
Thus,the solution is $x = 1, y = 1$.