Geodesic equation in terms of four velocity

I am trying to show that for timelike paths, we can write the geodesic equation in terms of the four-velocity $U^mu=frac{dx^mu}{dtau}$ as
$$U^lambdanabla_lambda U^mu=0.$$

In other words, substituting $U^ mu=frac{dx^mu}{dtau}$ into the above equation should produce the affinely parameterised geodesic equation

$$frac{d^2x^mu}{dtau^2}+Gamma^mu_{rholambda}frac{dx^rho}{dtau}frac{dx^lambda}{dtau}=0.$$

Performing the substitution, I got instead

$$U^lambda(partial_lambda U^mu +Gamma^mu_{rholambda}U^rho)=0$$
$$U^lambdapartial_lambda U^mu +Gamma^mu_{rholambda}U^rho U^lambda=0$$
$$frac{dx^lambda}{dtau} frac{partial}{partial x^lambda} frac{dx^mu}{dtau}+Gamma^mu_{rholambda}frac{dx^rho}{dtau}frac{dx^lambda}{dtau}=0$$
$$frac{dx^lambda}{dtau}frac{d}{dtau} frac{partial x^mu}{partial x^lambda}+Gamma^mu_{rholambda}frac{dx^rho}{dtau}frac{dx^lambda}{dtau}=0$$

$$frac{dx^lambda}{dtau}frac{d}{dtau} delta^mu_lambda+Gamma^mu_{rholambda}frac{dx^rho}{dtau}frac{dx^lambda}{dtau}=0$$

$$Gamma^mu_{rholambda}frac{dx^rho}{dtau}frac{dx^lambda}{dtau}=0$$

where I used $frac{d}{dtau} delta^mu_lambda=0$ in the last line since $delta^mu_lambda$ is a constant.

What am I doing wrong?