Does $X \rightarrow y$ being CDH imply that, given $X$ distinguishing between $y, r$ is DDH $\forall r\in Z_{p}^{*}$

In one proof I show that, given a cyclic group $Z_{p}^{*}$ where $p$ is prime, and a set of information $X$ computing $y\in Z_{p}^{*}$ is as difficult as solving Computational Diffie Hellman (CDH) problem.

In another proof can I make the argument that as transformation $X \rightarrow y$ is proven to be CDH, given $X$, distinguishing between any random $r\in Z_{p}^{*}$ from $y$ is as difficult as Decisional Diffie Hellman (DDH).

First, it might be that CDH is hard, yet the corresponding DDH is easy. In fact, this is generally the case in $\mathbb{Z}_p^*$, so you should be careful about that. See the footnote.

Second, there is a trivial counter example to your question: let $g$ be a generator of the subgroup of squares modulo $p$ (assume for simplicity that $p$ is a safe prime). Then, computing $g^{ab} \bmod p$ given $X = {g^a, g^b}$, for random $a,b$, is infeasible assuming the hardness of CDH over the subgroup of squares (which is implied by the hardness of CDH over $\mathbb{Z}_p^*$). Now, define $y$ as the value obtained by replacing the least significant bit of $g^{ab}\bmod p$ with zero. Computing $y$ from $X = {g^a, g^b}$ is as hard as CDH over the subgroup of squares (the reduction looses a factor 2). Yet, distinguishing $y$ from a random element is obviously not implied by DDH over this group (I focused on the subgroup of squares because DDH is broken over $\mathbb{Z}_p^*$, see below).

== Footnote ==

It is easy to distinguish $g^{ab} = y \bmod p$ from a random element $r$ of $\mathbb{Z}_p^*$: with probability 1/2, $r$ is not a square, and this is easy to check by computing the Legendre symbol (in contrast, $y$ is always a square).