$x$ , $y$ , and $z$

$\begin{cases} \dfrac {xy}{x+y} = 3 & \implies \dfrac {xy}3 & = x + y & \implies \dfrac {xyz}3 = zx + yz & ...(1) \\ \dfrac {yz}{y+z} = 4 & \implies \dfrac {yz}4 & = y + z & \implies \dfrac {xyz}4 = xy + zx & ...(2) \\ \dfrac {zx}{z+x} = 5 & \implies \dfrac {zx}5 & = z + x & \implies \dfrac {xyz}5 = yz + xy & ...(3) \end{cases}$

From $(1)+(2)+(3)$ :

$\begin{aligned} \left(\frac 13 + \frac 14 + \frac 15 \right)xyz & = 2(xy+yz+zx) \\ \implies \frac {xyz}{xy+yz+zx} & = \frac 2{\frac 13 + \frac 14 + \frac 15} = \frac {120}{47} \end{aligned}$

Therefore, $m+n = 120+47 = \boxed{167}$ .

The given system can be written as

$\dfrac{x + y}{xy} = \dfrac{1}{3} \Longrightarrow \dfrac{1}{y} + \dfrac{1}{x} = \dfrac{1}{3}$ $\dfrac{y + z}{yz} = \dfrac{1}{4} \Longrightarrow \dfrac{1}{z} + \dfrac{1}{y} = \dfrac{1}{4}$ $\dfrac{x + z}{xz} = \dfrac{1}{5} \Longrightarrow \dfrac{1}{z} + \dfrac{1}{x} = \dfrac{1}{5}$

Adding these together yields

$2 \left(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \right) = \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} = \dfrac{20 + 15 + 12}{60} = \dfrac{47}{60} \Longrightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{47}{120}$ .

So finally

$\dfrac{xyz}{xy + yz + zx} = \dfrac{1}{\dfrac{1}{z} + \dfrac{1}{x} + \dfrac{1}{y}} = \dfrac{1}{\dfrac{47}{120}} = \dfrac{120}{47} \Longrightarrow m + n = 120 + 47 = \boxed{167}$ .