Prove that $f:\mathbb{R}\to \mathbb{R}$ such that $$ f(x) = \left\{ \begin{array}{c l} x^2\, \sin\left(\frac{1}{x}\right) & ,\quad x\neq 0\\ 0 & ,\quad x=0 \end{array} \right.$$ is Lipschitz (without use of derivatives).
Attempt. I am aware (Lipschitz-continuous $f(x)=x^2\cdot \sin\left(\frac{1}{x}\right)$) that: $$|f(x)-f(y)|\leq 3|x-y| ~~~\forall~x,~y\in \mathbb{R},$$ but I am looking for a proof, without use of derivatives. I tried: for $x,y\neq 0$:
\begin{eqnarray} x^2\sin\frac{1}{x} - y^2 \sin\frac{1}{y} &=& (x^2-y^2)\sin\frac{1}{x} + y^2\left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right ),\nonumber \end{eqnarray} so: $$\left | x^2\sin\frac{1}{x} - y^2 \sin\frac{1}{y} \right | \leq |x^2 - y^2| + y^2 \left | \sin\frac{1}{x} - \sin\frac{1}{y} \right |.$$ Since: $$\left | \sin\frac{1}{x} - \sin\frac{1}{y} \right | \leq \left| \frac{1}{x} - \frac{1}{y}\right |= \frac{|x - y|}{xy},$$ we get: $$\left | x^2 \sin\frac{1}{x} - y^2 \sin\frac{1}{y} \right | \leq \left(x+y+\frac{y}{x}\right)|x-y|.$$ Unfortunatelly , the quantity $x+y+\frac{y}{x}$ grows to $+\infty$, either for large $x$, or for $x\approx 0.$
Thanks in advance for the help.