What's in a square root?

Square root of negative numbers are complex numbers, not undefined.

$\sqrt{x}$ will be undefined for negative $x$ if we are working in the field of real numbers, but it is of course defined in $\mathbb{C},$ the field of complex numbers. Then with $x = -a$ for real $a \gt 0,$ we find that

$\sqrt{x}*\sqrt{x} = \sqrt{-a}\sqrt{-a} = (i\sqrt{a})(i\sqrt{a}) = i^{2}\sqrt{a}\sqrt{a} = -a = x,$

where $i = \sqrt{-1}$ and thus $i^{2} = -1.$ So if we allow for complex numbers, then my solution method is valid for negative $x,y$ as well. If, however, $x,y$ are complex, i.e., have non-zero imaginary components, then this statement does not hold, since the square root function is not well-defined in $\mathbb{C}.$

(Note that the equation $\sqrt{x}*\sqrt{y} = \sqrt{xy}$ is only true in general for $x,y$ being non-negative reals.)

This issue is one of the most commonly disputed on this site. Every positive real $x$ does indeed have two square roots, one positive and one negative. (Of course $x = 0$ has a single square root.) However, in common usage, unless otherwise specified, when we are asked to evaluate $\sqrt{x}$ for $x \ge 0$ we are looking for the principal square root, which is the unique non-negative value $a$ such that $a^{2} = x.$

It gets a bit more complicated for negative reals, as we are then dealing with complex numbers, but as I've mentioned in a comment above my solution applies to negative reals as well.

If $x = y,$ then

$x - y = 0 \Longrightarrow (x - y)(x + y) = 0 \Longrightarrow x^{2} - y^{2} = 0 \Longrightarrow x^{2} = y^{2}.$

However, the reverse is not true. If $x^{2} = y^{2},$ then

$x^{2} - y^{2} = 0 \Longrightarrow (x - y)(x + y) = 0,$

in which case either $x - y = 0 \Longrightarrow x = y$ or $x + y = 0 \Longrightarrow x = -y.$

The reason I took the approach I did is because, in general, the "square both sides" approach can introduce extraneous solutions, so it's preferable to avoid using it if an alternative approach is available. As shown above, $x = y$ and $x^{2} = y^{2}$ are not equivalent, since the first equation has only one solution, namely $x = y,$ and the second equation has two solutions, namely $x = y$ and $x = -y.$

More specific to the stated problem, it can be implied that, since we are taking the square roots of $x$ and $y,$ these two variables are non-negative, and thus $\sqrt{x}$ and $\sqrt{y}$ are real and non-negative as well. In this case the "square both sides" approach is actually valid and will not yield extraneous solutions since the domain does not allow for it. That is, with the implicit assumption that $x$ and $y$ are non-negative reals, the solution sets of $\sqrt{x} = \sqrt{y}$ and $x = y$ are identical. However, my approach does not require this implicit assumption nor any discussion about extraneous solutions, so I felt that it was the best approach.

You don't think if a^(1/2) = b^(1/2), then a=b?

You have a point, but in common usage, unless otherwise specified, the square root function $f(x) = \sqrt{x}$ for $x \in \mathbb{R}^{+} \cup {0}$ returns the principal square root, which is unique and non-negative.

In your example we would have $x = (-2)^{2} = 4$ and $y = 2^{2} = 4,$ in which case $\sqrt{x} = \sqrt{y}$ does imply that $x = y.$

What your example does show is that $\sqrt{x^{2}} = \sqrt{y^{2}}$ does not necessarily imply that $x = y,$ but only that $|x| = |y|.$

By definition, over the non-negative reals the square root function returns a non-negative value, (i.e., the principal root).

The problem does not require a distinction, does it?

Taking $\sqrt{64}$ as a stand-in, the equation $x = y$ would become $64 = 64$ ; the values of the root are given as equal already, meaning they could be either (or both) roots, just as long as they were equal.

Any positive real number does indeed have two square roots, but unless specified otherwise, the function $f(x) = \sqrt{x}$ for $x \ge 0$ returns the non-negative principal square root. I suspect that this convention may not be universal, but I find that it holds true on this site and Wolfram, among others.

I've mentioned elsewhere in the comments that the given statement also holds true for negative reals, but that it does not hold true for $x,y \in \mathbb{C}$ , (with both of $x,y$ having non-zero imaginary components), as the square root function is not well-defined over the complex numbers.