Show that axiom of replacement implies from the Axiom of Specification [Proof Verification] 
Axiom $3.6$ (Replacement). Let $A$ be a set. For any object $x \in  A$ and any
  object $y$, suppose we have a statement $P(x, y)$ pertaining to $x$ and $y$,
  such that for each $x\in A$ there is at most  one $y$ for which $P(x,y)$ is
  true. Then there exists a set $\{y: P(x, y) \text{ is true for some } x \in A\}$ 
  such that for апy object $z$,  $$ z\in \{y : P(x, y)\text{ is true for some } x \in A\}  \iff  P(x,z)\text{ is true for some } x \in A.$$
Axiom $3.5$ (Specification). Let $A$ be a set, and for each  x$\in$  $A$, let $P(x)$ be a property pertaining to $x$ (i.e., $P(x)$ is either  a
  true statement or a false statement). Then there exists a set,  called
  $\{x \in A : P(x) \text{ is true}\}$ (or simply $\{x \in A : P(x) \text{ for short}\}$),  whose elements are precisely the elements $x$ in $A$ for which $P(x)$  is true. In other words, for any object $y$,  $$у \in \{x \in A: P(x)\text{ is true}\} \iff (y \in A \text{ and } P(y)\text{ is true}).$$

I have to show that $3.6\implies 3.5.$
Proof: By $(3.6)$ we can assume the following set $$\{x:P(x,x)\text{ is true for some }x\in A\}.$$ Let $Q(x)=P(x,x)$ then we get the set $$\{x\in A:Q(x) \text{ is true for some }x\},$$
which is what $(3.5)$ wants. Is this proof correct?
PS. I have read other answers to this question on MSE, but none of them use this formulation of the axiom and I guess use more formal notation. I am learning from Tao's Analysis book and so I've not been introduced to such notation. 

Your argument isn't quite correct. But it's on the way.
You are given a property $P$ and you want to separate a subset of $A$ which is specified by the property $P$.
For this, you need to find a property $Q(x,y)$ for which if $x\in A$, then there is at most a single $y$ such that $Q(x,y)$ holds, and you need to choose $Q$ in such a way that you get exactly the subset of $A$ specified by $P$.

The obvious solution, of course, is $Q(x,y)$ to be $x=y\land P(x)$. I will leave you to prove that it satisfies the conditions needed to apply Replacement on $Q$, and of course $\{y\mid Q(x,y)\text{ holds for some }x\in A\}$ is exactly—by the definition of $Q$—the same $\{x\in A\mid x=x\land P(x)\}$ which in turn is exactly $\{x\in A\mid P(x)\}$.
