We begin by moving the second inequality to the right side of the equation, giving us $ |x+5|=|3x-6| $. From here, we can split the equation into two separate cases. For the first case, note that if $ x+5 $ and $ 3x-6 $ have the same sign, then $ x+5=3x-6 $. Case 1: \begin{align*} x+5&=3x-6 \\\Rightarrow \qquad -2x&=-11 \\\Rightarrow \qquad x&=\frac{11}{2} \end{align*}If we plug this value of $ x $ back into the original equation to check our answer, we get that $ \left|\frac{11}{2}+5\right|-\left|3\left(\frac{11}{2}\right)-6\right|=0 $ or $ 0=0 $. Since this is true, we can accept $ x=\frac{11}{2} $ as a valid solution. For case two, note that if $ x+5 $ has a different sign than $ 3x-6 $, then $ x+5=-(3x-6) $. Case 2: \begin{align*} x+5&=-(3x-6) \\ x+5&=-3x+6 \\\Rightarrow \qquad 4x&=1 \\\Rightarrow \qquad x&=\frac{1}{4} \end{align*}If we plug this value of $ x $ back into the original equation to check our answer, we get that $ \left|\frac{1}{4}+5\right|-\left|3\left(\frac{1}{4}\right)-6\right|=0 $, which also gives us $ 0=0 $. This is always true, so we can accept $ x=\frac{1}{4} $ as a valid solution as well. Thus, our two possible solutions are $ \frac{1}{4} $ and $ \frac{11}{2} $. Since the question asks for the largest possible value of $ x $, our final solution is $ \boxed{\frac{11}{2}} $.