Square root of 9 from a technical perspective. When we say $$  \sqrt{9}= x $$
then $\;x = 3,\;$ right?
So why when we square both sides it becomes different:
$$ (\sqrt{9})^2 = x^2$$
$$9 = x^2$$
Here $\;x =\pm 3.$
So, does $\;x = \pm 3\;$ in $\sqrt{9} = x?$
 A: Just because $A$ is something that does $P$, does not mean that $A$ is the ONLY thing that does $P$.
It is true that $3^2 = 9$ but $3$ is not the only thing that when squared will equal $9$.
so $3^2 =9$ and $w^2 = 9$ does not in any way mean that $w = 3$.
This is no more valid or reasonable than saying $(1)^2- 3(1) +7 = 5$ and $(2^2 - 3(2) + 7 = 5$ and therefor $1 = 2$.
Or saying if Fred is a member of the chess club, and Betty is a member of the chess club, then Fred and Betty are the same person.
....
Your confusion is based on what is the definition of $\sqrt{9}$.
The definition of $\sqrt{k}$ has TWO parts.  If $\sqrt{k} = m$ then TWO things must be true:  1)  $m^2 = k$....  It's a common mistake to think this is the only thing that must be true but we must also have  2)  $m \ge 0$.
So although $3^2 = 9$ and $(-3)^2 = 9$ so both have 1) true.   $3>0$ but $-3< 0$ so only $3$ has 2) true and $-3$ has 2) false.
So $\sqrt 9 = 3$ and $-3\ne \sqrt 9$.
Now just because $3 = \sqrt 9$.  ANd $(\sqrt 9)^2 = 9$, it does not follow that $\sqrt 9$ is the ONLY thing that when squared equals $9$.  $\sqrt 9$ is only one thing that when squared equals $9$.
Just like Fred isn't the only member of the chess club and Betty can also be a member; $\sqrt 9$ is not the only thing that when squared is equal to $9$.  $-3$ when squared is also equal to $9$ when squared.
A: If $P$ implies $Q$, then this does not necessarily mean that $Q$ implies $P$. For example, all zebras are white-striped animals, but not all white-striped animals are zebras:
$$
\require{cancel}
\text{Zebra} \implies \text{Stripeyness}
$$
but
$$
\text{Stripeyness} \cancel{\implies} \text{Zebra} \, .
$$
The same principle applies here. We have
$$
x=3 \implies x^2=9
$$
but
$$
x^2 = 9 \cancel{\implies} x=3 \, . \\
$$
This is because every positive number has two square roots, and so just from the fact that $x^2=9$, we can't deduce that $x=3$. However, we certainly can deduce from $x=3$ that $x^2=9$.
A: Edited my answer.
As J. W. Tanner said in the comments, $\sqrt{9} = 3$. So $x = 3,\ $ not $-3$.
Therefore $x^2 = 9,\ $ because $3^2 = 9$.
Now the notation $t = \pm 3$ means that $\left(t=3\ \ or\ \ t = -3\right).$
So it is true that $x = \pm 3$ here because $\left(x=3\ \ or\ \ x = -3\right) $ is true.
Just because $x^2 = 9,\ $ doesn't mean that $x$ equals both $3$ and $-3$. It just means that $x = 3$ or $x = -3$ (or both). In this case, $x=3$ only, because $x$ was defined to be equal to $3$ only.
A: $a=b\implies a^2=b^2$ (merely by replacement of $b$ by $a$ in the right side of the implication, assuming the left side holds), but the converse is wrong in the sets $\mathbb R$, $\mathbb Z$, $\mathbb Q$, $\mathbb C$, and $\mathbb Z/p\mathbb Z$ for prime $p>2$. Proof by example: $a=1$, $b=-1$.
However, $a^2=b^\implies a=b$ in some other sets, including $\mathbb R_+$, $\mathbb N$, and the Booleans $\mathbb Z/2\mathbb Z=\{0,1\}$.

So, does $\;x = \pm 3\;$ in $\sqrt{9} = x$ ?

No in $\mathbb N$, $\mathbb Z$, $\mathbb Q$, $\mathbb R$, where a square root is non-negative by definition. Yes in sets where $-3=3$; or for some unusual definition of square root.
