How to prove using $\epsilon - \delta$ that $\lim_{(x,y)\rightarrow (1,0)}{\frac{(x-1)^2 \ln x}{y^2 +(x-1)^2}} =0$

Question

I am stuck on this homework question.I know the "regular" methods to evaluate this limit but I have no clue how to prove that this limit's value is $0$ using $\epsilon - \delta$. I have tried using circular neighborhood, i.e, using $0<\sqrt{(x-1)^2+y^2}<\delta$, as well as using square neighborhood, i.e., using $0<|x-1|<\delta $ and $0<|y|<\delta$, but nothing seems to work. Please note that I can use only $\epsilon - \delta$ approach. Using Squeeze Theorem is not allowed. I just need a hint.... proper answer is not required. Thanks in advance :).

Answer 1

By using @metamorphy hint:

$|\frac{(x-1)^2\ln{x}}{y^2 + (x-1)^2}| = |\frac{(x-1)^2}{y^2 + (x-1)^2}| |\ln{x}| \leqslant |\ln{x}|$

So it suffices to show it for 1 variable function $\ln{x}$ as $x\rightarrow 1$.