How can I prove that $f(x,y)=x^2|y|$ satisfies Lipschitz condition?

Question

on the rectangle $|x| \le 1$, $y \le 1$?

My book says that $\frac{f(x,y_1)-f(x,y_2)}{y_1-y_1}$ must be bounded.

I got that $\frac{f(x,y_1)-f(x,y_2)}{y_1-y_1}= \frac{x^2(|y_1|-|y_1|)}{y_1-y_2}$ but I don't know how to proceed since if $y_1=y_2$ then this quotient is not defined.

Answer 1

$x^2 \leq 1$, and $\dfrac{|y_1|-|y_2|}{|y_1-y_2|} \leq 1\implies \left|\dfrac{f(x,y_1)-f(x,y_2)}{y_1-y_2}\right| \leq 1$