I think factorization may help

$\frac { x^{ 2 }-1 }{ 2x } +\frac { x^{ 3 }-1 }{ 3x } +\frac { x^{ 4 }-1 }{ 4x } =0$

Multiply x to all terms.

$\frac { x^{ 2 }-1 }{ 2 } +\frac { x^{ 3 }-1 }{ 3 } +\frac { x^{ 4 }-1 }{ 4 } =0$

Split all fractions.

$\frac { x^{ 2 } }{ 2 } -\frac { 1 }{ 2 } +\frac { x^{ 3 } }{ 3 } -\frac { 1 }{ 3 } +\frac { x^{ 4 } }{ 4 } -\frac { 1 }{ 4 } =0$

Simplifying,

$\frac { 3x^{ 4 }+4x^{ 3 }+6x^{ 2 }-13 }{ 12 } =0$

Multiply 12 both sides.

$3x^{ 4 }+4x^{ 3 }+6x^{ 2 }-13=0$

By Vieta's Formula we get sum of roots equal to $\frac { -b }{ a }=\frac { -4 }{ 3 }$

So mean is

$\frac { \frac { -4 }{ 3 } }{ 4 } =\boxed { \frac { -1 }{ 3 } }$

Factorise each of the fractions as follows: $(x-1) [\frac{x+1}{2x}+\frac{x^2 -x +1}{3x} + \frac{(x+1)(x^2 +1)}{4x} ]= 0$ Now make the denominator 12x for all terms: $(x-1)[\frac{(6x+6) + (4x^2 -4x+4) +(3x^3 +3x^2 +3x +1)}{12x} ] = 0$ $(x-1) [\frac{3x^3 +7x^2 +5x +13}{12x} ] = 0$ Now using Vieta root sums: $\sum_{}^{} roots = 1-\frac{7}{3} = -\frac{4}{3}$ $\therefore\ AM = -\frac{\frac{4}{3} }{4} = -\frac{1}{3}$