No woods or log's required!

Since we get 3 equations, we can say that $27^{x} = 60, 64^{y} = 60, 125^{z} = 60$ .

Take a x,y,z-th root we get...

$27 = 60^{\frac{1}{x}}, 64 = 60^{\frac{1}{y}}, 125 = 60^{\frac{1}{z}}$ .

We multiply 3 equations we get...

$27\times64\times125\ = 60^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$

$60^{3} = 60^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 3$ ....

Simplify that we get $\frac{xy+yz+zx}{xyz} = 3$

$\frac{xyz}{xy+yz+zx} = \frac{1}{3}$

Therefore: $\frac{2013xyz}{xy+yz+zx} = \frac{2013}{3}\ = \boxed{671}$ ~~~

${ 27 }^{ x }={ 64 }^{ y }={ 125 }^{ z }=60\\ \therefore \quad x=\log _{ 27 }{ 60 } ,\quad y=\log _{ 64 }{ 60 } ,\quad z=\log _{ 125 }{ 60 } \\ Now,\frac { 2013xyz }{ xy+yz+zx } =\quad \frac { 2013 }{ \frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } } \quad (Dividing\quad numerator\quad and\quad denominator\quad by\quad xyz)\\ \therefore \frac { 2013xyz }{ xy+yz+zx } =\quad \frac { 2013 }{ \frac { 1 }{ \log _{ 27 }{ 60 } } +\frac { 1 }{ \log _{ 64 }{ 60 } } +\frac { 1 }{ \log _{ 125 }{ 60 } } } \\ \qquad \qquad \qquad \quad =\quad \frac { 2013 }{ \log _{ 60 }{ 27 } +\log _{ 60 }{ 64 } +\log _{ 60 }{ 125 } } \\ \qquad \qquad \qquad \quad =\quad \frac { 2013 }{ 3\log _{ 60 }{ (3\times 4\times 5) } } =\frac { 2013 }{ 3\log _{ 60 }{ (60) } } =\frac { 2013 }{ 3 } =\boxed { 671 }$

Firstly, we have no direction on how to solve this problem, we can either handle

$27^x=64^y=125^z=60$

or

$\dfrac{2013xyz}{xy+yz+xz}$

We cannot really see the connection between these two expressions and the first equation seems complicated if we touch it. So, what I did is simplify the second expression:

$\dfrac{2013xyz}{xy+yz+xz}$

$=2013\Big(\dfrac{1}{\frac{xy+yz+xz}{xyz}}\Big)$

$=2013\Big(\dfrac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\Big)$

Now, the expression become more simpler, the connection between them is as below

1. $60^\frac{1}{x}=27$

2. $60^\frac{1}{y}=64$

3. $60^\frac{1}{z}=125$

Then, we do as below:

$60^{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=3^3\times4^3\times5^3=60^3$

So,

${\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=3$

Substitute this to the given expression we have:

$=2013\bigg(\dfrac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\bigg)$

$=2013(\frac{1}{3})$

$=671$

$27^{x} = 3^{3x} = 64^{y} = 2^{6y} = 125 ^{ z}$ = $5 ^ {3z} = 60$ and on multiplying 3 equations we get $[ 3 ^ {x} * (2 ^ {2}) ^ {y} * 5 ^ {z}] ^ {3} = 60 ^ {3}$ and further we get $3 ^{ x }* 2 ^ {2}y * 5 ^ {z} = 3 ^ {1} * 2 ^ {2} * 5 ^ {1}$ and hence we get x = y = z = 1 and then $\frac{2013 * x * y * z} {(xy + yz + xz)}$ = $\frac{2013}{3 }= \boxed{671}$