In this excercise I want to show that $\nabla_\rho g_{\mu \nu}=0$ and $\nabla_\rho g^{\mu \nu}=0$
This should probably be very easy, but excuse me I'm completly new to GR.
So to do this I used that the covariant derivative of a rank 2 tensor is
$$ \nabla_c t^{ab} = \partial_c t^{ab} + \Gamma^a_{dc}t^{db}+\Gamma^b_{dc}t^{ad} $$ $$ \nabla_c t_{ab} = \partial_c t_{ab} - \Gamma^d_{ac}t_{db}-\Gamma^d_{bc}t_{ad} $$
And I rewrote the connection in therms of partial derivatives of the metric. So for the covariant components of the metric this worked out nicely, everything canceld and ended up to be zero.
Now with the contravariant components I'm running into trouble:
$$ \nabla_\rho g^{\mu \nu} = \partial_\rho g^{\mu\nu} + \Gamma^\nu_{d\rho}g^{d\nu}+\Gamma^v_{d\rho}g^{\mu d}=\\ \partial_\rho g^{\mu\nu} + \frac{1}{2}g^{\mu e} (\partial_d g_{e \rho} + \partial_\rho g_{de} -\partial_e g_{d\rho})g^{d \nu}+ \frac{1}{2}g^{\nu e} (\partial_d g_{e \rho} + \partial_\rho g_{de} -\partial_e g_{d\rho})g^{\mu d} $$
Now the two parts inside the brakets are the same and the metrics outside replace the dummy indices d and e by $\nu$ and $\mu$. If I raie the indices d and e four of the six terms cancel each other but then I am left with:
$$ \nabla_\rho g^{\mu \nu} = 2 \partial_\rho g^{\mu \nu} $$
This is clearly not zero for a general metric in non local coordinates. So I assume I'm messing something up completly, pleas ehelp me to understand this.