Is this contraction of metric tensor derivatives symmetric?

Since the product rule tells us $0 = \partial( g g^{-1} ) = (\partial g) g^{-1} + g (\partial g^{-1})$, we have a formula for the derivative of the inverse metric:

$$ \partial_l g^{ij} = -g^{ia} g^{jb} \partial_l g_{ab}.$$

Substituting this in to your expression we get

$$ -g^{ia} g^{jb} \partial_l g_{ab} \partial_k g_{ij}.$$

If we swap the dummy indices $a \leftrightarrow i$, $b \leftrightarrow j$ then this is equal to

$$ -g^{ai} g^{jb} \partial_l g_{ij} \partial_k g_{ab};$$

so it's symmetric in $k$ and $l$.

It does appear to be symmetric, though the proof I came up with requires the introduction of a covariant derivative operator. There may be another proof out there that doesn't require quite so much heavy machinery.

Let $\nabla_k$ be a torsion-free derivative operator defined such that $\nabla_k g_{ij} = 0$. By the general properties of derivative operators, we know that there will exist a tensor $C^i {}_{jk}$ such that the coordinate derivative and our new derivative of a covariant rank-2 tensor are related by $$ \nabla_k g_{ij} = \partial_k g_{ij} - C^m {}_{ki} g_{mj} - C^m {}_{kj} g_{im} $$ and similarly, the derivative of a contravariant rank-2 tensor are given by $$ \nabla_k g^{ij} = \partial_k g^{ij} + C^i {}_{km} g^{mj} + C^j {}_{km} g^{im}. $$ Since by definition $\nabla_k g_{ij} = 0$ and $\nabla_l g^{ij} = 0$, the quantity in your question becomes $$ \partial_k g_{ij} \partial_l g^{ij} = (C^m {}_{ki} g_{mj} + C^m {}_{kj} g_{im})(-C^i {}_{ln} g^{nj} - C^j {}_{ln} g^{in}) = - 2 C^m {}_{ki} C^i {}_{lm} - 2 C^m {}_k {}^j C_{mlj}. $$ Both terms in this expression are manifestly symmetric under the exchange of $k$ and $l$, and so the expression $\partial_k g_{ij} \partial_l g^{ij}$ is symmetric as well.