Metric Derivative Proof

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How do I go about proving

$\partial_{\mu} g^{\nu \rho}=-g^{\nu \sigma}g^{\rho \lambda}\partial_{\mu} g_{\sigma \lambda}$?

I've tried using the covariant derivative and the Christoffel symbols but it seems to be to no end in finding this specific relation.

Any ideas?

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(Just to make this question answered)

Observe that for any invertible matrix

$$ A^{-1} A = I $$

where $I$ is the identity matrix. Now, let $A(t)$ be a parametrised family of invertible matrices, you have by the product rule:

$$ (\partial_t A^{-1}) A + A^{-1} \partial_t A = 0 $$

since the identity matrix is independent of $t$. Therefore

$$ (\partial_t A^{-1}) A = - A^{-1} \partial_t A \implies \partial_t A^{-1} = - A^{-1} (\partial_t A ) A^{-1} $$

writing out the matrix multiplication in index notation you get exactly the expression you wanted.