Given: $y=\frac{x}{3(x^2-3x5)}$; Find $y^{\text{'}}.$

$\frac{i}{2}[\frac{5-x}{(x^2-3x5{)}^2}]$

$\frac{1}{3}[\frac{5-x^2}{(x^2-3x5{)}^2}]$

$\frac{x^2-5}{(x^2-3x5{)}^2}$

$\frac{1}{3}[\frac{x^2-5}{(x^2-3x5)}]$

$($E$)$ None of the choices