Functional Equation: $f(x^2-y^2)=xf(x)-yf(y)$ 
Let $\mathbb{R}$ be the set of Real numbers. Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that $$f(x^2-y^2)=xf(x)-yf(y)$$ for all pairs of real numbers $x$ and $y$.

This is a problem from USAMO 2002. The solution on the page I linked, is a bit tedious (maybe for good reason). I tried it this way. Is it valid?

Putting $x=y$ yields $f(0)=0$. 
Putting $x=-y$ gives $$0=f(0)=xf(x)+xf(-x)=x(f(x)+f(-x))$$ which is true for all $x\in\mathbb{R}$ and hence, $f(-x)=-f(x)$ and the function is odd. Since the function is odd, I'll consider only $x\in\mathbb{R}_+$
Now, this is the part I'm not sure about.
$$f(x^2)=xf(x)=x\sqrt{x}f(\sqrt{x})=\dots=\lim_{n\to\infty}x^{1+\frac{1}{2}+\frac{1}{4}+\dots+\frac{1}{2n}}f(x^\frac{1}{2n})=x^2\lim_{n\to\infty}f(x^\frac{1}{2n})$$
For non-zero $x$, $$\lim_{n\to\infty}x^\frac{1}{2n}=1$$ (How do I prove this? I've seen it work on a calculator, and I know that for $x\in(0,1)$, $x^{\frac{1}{n}}>x$ and for $x\in(1,\infty)$, $x^{\frac{1}{n}}<x$ and that $1$ is a fixed point. Is that sufficient?)
So, $$f(x^2)=x^2f(1)=cx^2$$ and with $x^2=t$ we arrive at $$f(t)=ct$$ This also satisfies the odd criterion.

Is the proof above correct? Am I assuming somethings without realising? I know these are risky waters, so I'd rather be careful.
EDIT: I realise that I'm assuming continuity when I take the limit inside: $$\lim_{n\to\infty}f(x^\frac{1}{2n})=f(\lim_{n\to\infty}x^\frac{1}{2n})$$ So, the naturally next question is how can I prove continuity?
 A: As the solution on the page you linked notes, all solutions must be additive: 
$$ f(x+y) = f(x) + f(y)$$
Now there are discontinuous additive functions.  But in this case we have additional structure.  In fact, taking $x = s+1$ and $y = s-1$ in your original equation, you can simplify the result (using additivity) to get 
$$ f(s) = s f(1)$$
A: Let
$$f(x^2−y^2)=xf(x)−yf(y) \implies P(x,y)$$
$$P(0,0) \implies f(0)=0$$
$$P(x,0) \implies f(x^2)=xf(x) \tag{1}$$
$$P(-x,0) \implies f(x^2)=-xf(-x) $$
and substituting that in $(1)$ we get 
$$f(-x)=-f(x) \tag{2}$$
$$P(x,y) \implies f(x^2−y^2)=xf(x)−yf(y)$$
$$ \Leftrightarrow f(x^2−y^2)=f(x^2)−f(y^2) \tag{by (1)}$$
$$ \Leftrightarrow f(x^2−y^2)=f(x^2)+f(-y^2) \tag{by (2)}$$
Substituting $x^2=u$ and $-y^2=v$ we get
$$f(u+v)=f(u)+f(v) \implies H(u,v)$$
(Note that my substitution works if $u \geq 0$ and $v \leq 0$, but by $(2)$, $f$ is odd, so that extends $H(u,v)$ to all $u,v \in \mathbb{R}$)
At this point, we can partition any sum inside a function as we like:
$$f(a_1+a_2+\dots+a_n)=f(a_1)+f(a_2)+\dots+f(a_n) \tag{3}$$ 
for a positive integer $n$ and variables $a_i$
$$H((t+1)^2,0) \implies f(t^2+2t+1)=f((t+1)^2)$$
$$\Leftrightarrow f(t^2)+f(2t)+f(1)=(t+1)f(t+1) \tag{by (1) and (2)}$$
$$\Leftrightarrow tf(t)+f(t+t)+f(1)=(t+1)(f(t)+f(1)) \tag{by (1) and (2)}$$
$$\Leftrightarrow tf(t)+f(t)+f(t)+f(1)=tf(t)+tf(1)+f(t)+f(1) \tag{by (2)}$$
$$\Leftrightarrow f(t)=tf(1)$$
Putting $f(1)=k$ and substituting in the original equation, we get that $f(x)=kx$ is valid for all $x,k \in \mathbb{R}$
