Using the Lagrange Multipliers Method to prove $\ |h(x,y)|\leq 1$ 
Using the Lagrange multipliers method, I have found that the  maximum and minimum values of the function 
  $$f(x,y)=xy$$
  on the curve
  $$x^2-yx+y^2=1$$ are
  $$1 \ \text{at} \ (\pm1,\pm1) \ \ \ \text{and} \ \ -\frac{1}{3} \ \text{at} \ \big(\pm\frac{1}{\sqrt{3}},\mp\frac{1}{\sqrt{3}}\big) \ \ \text{respectively.}$$ 
  Using this, prove that $$\Big|\frac{xy}{x^2-yx+y^2}\Big|\leq1 \ \ \ \forall(x,y)\neq0$$

I don't really know where to start. I thought of multiplying across as the inequality will be preserved, which yields
$$|xy|\leq|x^2-yx+y^2|$$
then
$$|xy|-|x^2-yx+y^2|\leq 0$$
At this point, I'm relying on my algebraic manipulation skills to yield something true. I'm not using the information above obtained by the Lagrange multipliers method.

Equivalently, you prove: $(xy)^2 - (x^2-xy+y^2)^2 \le 0\iff (xy-x^2+xy-y^2)(xy+x^2-xy+y^2) \le 0\iff(2xy-x^2-y^2)(x^2+y^2) \le 0\iff -(x-y)^2(x^2+y^2) \le 0$ which is true.
We can show the max/min result without bothering Lagrange ! We have: 
$(x-y)^2 \ge 0 \implies x^2-2xy+y^2 \ge 0 \implies x^2 -xy+y^2 \ge xy \implies 1 \ge xy\implies \text{max}  = 1$
And observe $xy \ge -\dfrac{x^2-xy+y^2}{3}= -\dfrac{1}{3}\implies \text{min} = -\dfrac{1}{3}$ .
