Difficulty in finding appropriate $\delta$ I'm trying to prove that for every $\epsilon > 0$ there exists $\delta = $___
s.t for every $0 < \lvert\lvert(x,y) - (1, 1)\rvert\rvert < \delta$ :
$\lvert x^2y\rvert+\lvert y \rvert + 1 < \epsilon$

Can you help me find such $\delta$?
This is what I got so far:
$\lvert x^2y\rvert+\lvert y \rvert + 1 = \lvert y \rvert ( \lvert x \rvert ^2 + 1) + 1 \leq (\delta + 1)( \lvert x \rvert ^2 + 1) +1$
($\lvert y-1 \rvert = \sqrt{(y-1)^2} \leq \sqrt{(x-1)^2 + (y-1)^2} = \lvert\lvert(x,y) - (1, 1)\rvert\rvert < \delta$)
($\lvert y-1 \rvert < \delta \implies \lvert y \rvert < \delta + 1$)

I suspect you are looking at the function from two variables to two given by
$$ g(x,y) = (x^2 y,y) $$ so that $g(1,1) = (1,1).$ Then continuity of $g$ would be the ability to find a $\delta = \delta(\varepsilon)$ such that, when $  \| (x,y) - (1,1)  \| < \delta,$ then
$$  \| g(x,y) - g(1,1) \| < \varepsilon,  $$ or
$$  \| g(x,y) - (1,1) \| < \varepsilon,  $$ or 
$$  \| (x^2 y - 1,y - 1)  \| < \varepsilon  $$
