¿$Cov(X,Y) = 1$ and $Cov(Y,Z) = 1$ implies $Cov(X,Z) = 1$?

In general it is not true. Consider the following example: Let $Y=X+Z\implies Cov(X,Y)=Cov(X,X+Z)=Var(X)+Cov(X,Z)$, also $Cov(Y,Z)=Cov(X+Z,Z)=Cov(X,Z)+Var(Z)$. We wish to have the condition that $Cov(X,Y)=Cov(Y,Z)=1$ $$\implies Var(X)+Cov(X,Z)=Cov(X,Z)+Var(Z)\implies Var(X)=Var(Z)=1$$ We suppose that the variances are equal to 1 to illustrate the example. And from $Cov(Y,Z)=Cov(X,Z)+Var(Z)=1$ it follows that $Cov(X,Z)=1-Var(Z)=1-1=0$. So, even without any assumptions on distributions we get that $Cov(X,Y)=Cov(Y,Z)=1$ implies $Cov(X,Z)=0$. This is one possible counterexample.