As Deane pointed out the formula in
MathWorld looks wrong.
They use the following derivation
\begin{align}
{A_{\mu\nu\,;\,\lambda}}^{;\,\lambda}
&=\Big(g^{\lambda\kappa}A_{\mu\nu\,;\,\lambda}\Big)_{;\,\kappa}\tag{1}\\
&=g^{\lambda\kappa}
\frac{\partial^2A_{\mu\nu}}{\partial x^\lambda\partial x^\kappa}
-g^{\color{red}{\mu\nu}}\,{\Gamma^\lambda}_{\color{red}{\mu\nu}}
\frac{\partial A_{\color{red}{\mu\nu}}}{\partial x^\lambda}\tag{2}\\
&\dots\\
&=\frac{1}{\sqrt{g}}\Big(\sqrt{g}\,g^{\mu\kappa}\,A_{\mu\nu,\,\kappa}\Big)_{,\,\mu}\,\tag{5}
\end{align}
where $A_{\mu\nu\,;\,\lambda}$ is the covariant derivative, $g_{\mu\nu}$ is the
metric tensor, $g$ its determinant and $A_{\mu\nu\,,\,\kappa}$ is the ordinary partial derivative.
The term (2) has obvious errors that I highlighted in red. Summation
convention doubles indices and does not triple them, and the indices
$\mu\nu$ cannot be summed away completely.
Even if we try a fix with some dummy indices
$$
g^{\color{blue}{\alpha\beta}}\,{\Gamma^\lambda}_{\color{blue}{\alpha\beta}}
\frac{\partial A_{\color{green}{\mu\nu}}}{\partial x^\lambda}\tag{2'}
$$
it looks not convincing as I will show below.
The
whole derivation and the formula (5) are highly questionable.
A correct derivation should start as follows and it will soon become clear that
things are not simple:
First,
\begin{align}
A_{\mu\nu\,;\,\lambda}=A_{\mu\nu\,,\,\lambda}
\underbrace{-{\Gamma^\rho}_{\mu\lambda}A_{\rho\nu}}_{(*)}\;
\underbrace{-{\Gamma^\rho}_{\nu\lambda}A_{\mu\rho}}_{(**)}\,.\tag{A}
\end{align}
Since the Christoffel connection is metric compatible we have ${g ^{\lambda\kappa}}_{;\,\alpha}=0$ so that
$g^{\lambda\kappa}$ can be pulled out of a covariant derivative as if it were a constant. This leads to
\begin{align*}
{A_{\mu\nu\,;\,\lambda}}^{;\,\lambda}&=g^{\lambda\kappa}\,A_{\mu\nu\,;\,\lambda\kappa}\\
&=g^{\lambda\kappa}\Big\{A_{\mu\nu\,;\,\lambda,\,\kappa}-{\Gamma^\rho}_{\mu\kappa}A_{\rho\nu\,;\,\lambda}
-{\Gamma^\rho}_{\nu\kappa}A_{\mu\rho\,;\,\lambda}-{\Gamma^\rho}_{\lambda\kappa}A_{\mu\nu\,;\,\rho}\Big\}\,.
\end{align*}
Plugging in the three terms for each covariant derivative of $A_{\alpha\beta}$
from (A) leads to
a total of, I believe, fourteen terms which contain partial derivatives of Christoffel symbols
because the product rule needs to be applied on (*) and (**).
