How to show that two normally distributed density functions are independent Show that $U$ and $V$ are independent, where $U=X+Z$ and $V=X-Z$.

I´m given that $X\sim N(0,1)$, $Z\sim N(0,1)$ and X and Z are independent.

First I find U and V.
This leads to $U\sim N(0,2)$ and $V\sim(0,2)$

From here I'm stuck, I know that $independence \implies Cov(V,U)=0$ but i'm not sure if $Cov(V,U)=0 \implies independence.$
The suggestion in the comment is correct. If you can show that U and V are jointly gaussian with covariance zero then you have their independence. There are details about this lemma online, for example this page.

In this case, as X and Z are independent it's easy to show that all linear combinations of U and V are gaussian RVs. From there you can show the independence.

You are also correct that zero covariance does not imply independence. For example, $A$ is a RV with $\mathbb{E}A =  0$ and $\mathbb{E}[A^2] = 0$, and let $B=A$. Then we have $Cov(AB) = 0$ but $A$ and $B$ are clearly dependent.
The suggestion in the comment is correct. If you can show that U and V are jointly gaussian with covariance zero then you have their independence. There are details about this lemma online, for example this page.

In this case, as X and Z are independent it's easy to show that all linear combinations of U and V are gaussian RVs. From there you can show the independence.

You are also correct that zero covariance does not imply independence. For example, $A$ is a RV with $\mathbb{E}A =  0$ and $\mathbb{E}[A^2] = 0$, and let $B=A$. Then we have $Cov(AB) = 0$ but $A$ and $B$ are clearly dependent.