Does $f\big(x^2-y^2\big)=x\cdot f(x)-y\cdot f(y)$ imply $f\big(x^2\big)=x^2\cdot f(1)$?

A function $f$ is defined over reals and $f\left(x^2 - y^2\right) = x\cdot f(x) - y \cdot f(y)$ holds for all $x$ and $y$.

Plugging in $y = 0$ we get: $$f\left(x^2\right) = x \cdot f(x)$$ Let $x>0$: $$f\left(x^2\right) = x\cdot f(x) = x \cdot x^\frac{1}{2} \cdot f\left(x ^ \frac{1}{2}\right) = \dots = x^{1 + \frac{1}{2} + \dots +\frac{1}{2^n}} \cdot f\left(x^{\frac{1}{2^n}}\right) = x^{2 - \frac{1}{2^n}} \cdot f\left(x^{\frac{1}{2^n}}\right) \text,$$ for all positive integers $n$.

Does this imply $f\left(x^2\right) = x^{2-0} \cdot f\left(x^0\right) = x^2 \cdot f(1)$?

Your argument is sound assuming that $ f $ is continuous at $ 1 $. But in fact, you don't need to assume continuity at any point to find the solution.

Assuming $ f : \mathbb R \to \mathbb R $ satisfies $$ f \left( x ^ 2 - y ^ 2 \right) = x f ( x ) - y f ( y ) \text , \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $, you can let $ y = 0 $ in \eqref{0} to get $$ f \left( x ^ 2 \right) = x f ( x ) \text . \tag 1 \label 1 $$ You can use \eqref{1} to rewrite the right-hand side of \eqref{0}, which shows $ f ( x - y ) = f ( x ) - f ( y ) $ for all nonnegative $ x $ and $ y $. Equivalently, we get $ f ( x + y ) = f ( x ) + f ( y ) $ for all $ x $ and $ y $ such that $ y \ge 0 $ and $ x + y \ge 0 $. As for any $ x , y \in \mathbb R $ we know that $ | x | $, $ | y | $, $ | x | + | y | $, $ x + | x | $, $ y + | y | $ and $ x + y + | x | + | y | $ are all nonnegative, we have $$ f ( x + y ) = f \big( ( x + y + | x | + | y | ) - ( | x | + | y | ) \big) \\ = f ( x + y + | x | + | y | ) - f ( | x | + | y | ) \\ = \big( f ( x + | x | ) + f ( y + | y | ) \big) - \big( f ( | x | ) + f ( | y | ) \big) \\ = \big( f ( x + | x | ) - f ( | x | ) \big) + \big( f ( y + | y | ) - f ( | y | ) \big) \text , $$ and finally $$ f ( x + y ) = f ( x ) + f ( y ) \text . \tag 2 \label 2 $$ Now, we can let $ a = f ( 1 ) $ and use \eqref{1} and \eqref{2} to get $$ ( x + 1 ) \big( f ( x ) + a \big) = ( x + 1 ) f ( x + 1 ) = f \left( ( x + 1 ) ^ 2 \right) = \\ f \left( x ^ 2 + x + x + 1 \right) = f \left( x ^ 2 \right) + 2 f ( x ) + a = ( x + 2 ) f ( x ) + a \text , $$ which can be simplified to $ f ( x ) = a x $. It's straightforward to check that every function of this form indeed satisfies \eqref{0}, and thus we've found all the solutions.