Why is the domain of $x^2$ the set of all real numbers? I try to understand, why domain of $x^2$ is the set of all real numbers.
My doubts:
The domain of square root is not defined for negative numbers. Reason to that (If I am not wrong) is that function is supposed to have output with only one input leading to it. Therefore 9 cannot be square-rooted to -3 and 3, only to the positive number (in this case it's 3). Is not the situation with  $x^2$ the same? Should not a domain for it be limited to non-negative numbers? Otherwise, we are having an ambiguity of 3 and -3 leading to 9...

You're right about one thing: the square root function is the inverse function to the squaring function, after the squaring function is restricted to a domain on which it's one-to-one.  That domain is nonnegative numbers.
However, I think you're overthinking this problem.  When a problem asks for the the domain of a function defined by an algebraic expression, the task is to calculate the entire subset of real numbers which can be substituted into the expression.  For instance, if the expression is $\frac{1}{1-x^2}$, you're supposed to notice that the denominator cannot be zero for this to make sense. So the domain must carve out any numbers which do that, namely $\pm 1$.  Therefore the domain is $\mathbb{R}\setminus\{-1,1\}$.
But the expression you're given is just $x^2$.  This is defined for all real numbers.  So the domain is $\mathbb{R}$.
