Nakahara says on page 279 that the Killing equation
$$ X^\xi\partial_\xi g_{\mu\nu}+\partial_\mu X^\kappa g_{\kappa\nu}+\partial_\nu X^\lambda g_{\mu\lambda}=0 $$
can be written as
$$ (\mathcal L_X g)_{\mu\nu}=0 $$
Why are these two equations equal? I tried to use the definition of the Lie derivative like this:
$$ (\mathcal L_X g)(Y,Z) = \mathcal L_X(g(Y,Z)) - g(\mathcal L_XY,Z)-g(Y,\mathcal L_XZ) $$
but I can't see how I can use this to prove their equality.
I'm not sure where you're stuck, but $\mathcal L_X g$ is a $(0,2)$ tensor, with components given by
$(\mathcal L_X g)_{\mu\nu} := (\mathcal L_X g)(\partial_\mu, \partial_\nu) = \mathcal L_X(g(\partial_\mu,\partial_\nu)) - g(\mathcal L_X \partial_\mu,\partial_\nu)-g(\partial_\mu,\mathcal L_X \partial_\nu).$
Then you can use the fact that
$\mathcal L_X \partial_\mu = [X,\partial_\mu] = -[\partial_\mu, X^\kappa\partial_\kappa] = X^{\kappa}[\partial_\mu,\partial_\kappa] -\partial_\mu(X^{\kappa})\partial_\kappa =-\partial_\mu(X^{\kappa})\partial_\kappa$
to get
$ - g( \mathcal L_X \partial_\mu , \partial_\nu ) = - g( - \partial_\mu(X^{\kappa})\partial_\kappa, \partial_\nu )= \partial_\mu X^{\kappa}g_{\kappa\nu}.$
Use the same scheme on the last term.
For the first term, use the fact that $\mathcal L_X (f) = Xf$ for a vector field $X$ and function $f$ (the argument $g(\partial_\mu,\partial_\nu) = g_{\mu\nu}$ is just a function).