<s>[INST] 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:
$\lvert x^2y\rvert+\lvert y \rvert + 1 = \lvert y \rvert ( \lvert x \rvert ^2 + 1) + 1 \leq (\delta + 1)( \lvert x \rvert ^2 + 1) +1$
($\lvert y-1 \rvert = \sqrt{(y-1)^2} \leq \sqrt{(x-1)^2 + (y-1)^2} = \lvert\lvert(x,y) - (1, 1)\rvert\rvert < \delta$)
($\lvert y-1 \rvert < \delta \implies \lvert y \rvert < \delta + 1$)
 [/INST] I'm going to assume this is related to this question Is $f$ continuous at $(0, 1)$ , $(1,1)$?.
The similarities are too great.
So that question was to show 
$f(x,y) =
\left\{
 \begin{array}{ll}
  x^2y  & \mbox{if } x \in \mathbb{Q} \\
  y & \mbox{if } x \notin \mathbb{Q}
 \end{array}
\right.$
show that $f$ is continuous at $(1,1)$
So you want to show for any $\epsilon > 0$ there is a $\delta$ such that
$||(x,y) - (1,1)|| < \delta \implies |f(x,y) - 1| < \delta$.
==== argh! here is a much better answer =====
Let $0 < 1/n < \epsilon/4$
Let $\delta = 1/n$.
If $|(x,y) - (1,1)| < 1/n$ then $(n-1)/n< y < (n+1)/n \implies |y -1| < 1/n < \epsilon$.
Likewise $(n-1)/n< x < (n+1)/n$ so $[(n-1)/n]^3 < x^2y < [(n+1)/n]^3$ so
$1 - 3/n + 3/n^2 - 1/n^3 = 1- 1/n[3-3/n+1/n^2] < x^2y < 1 + 3/n + 3/n^2 + 1/n^3= 1 + 1/n[3+3/n+1/n^2]$
$1 - 4/n < 1- 1/n[3-3/n+1/n^2] < x^2y < 1 + 3/n + 3/n^2 + 1/n^3= 1 + 1/n[3+3/n+1/n^2] < 1 + 4/n$
So $|x^2y - 1| < 4/n < \epsilon$.
So as $f(x,y) = x^2y$ or $f(x,y) = y$.  $|f(x,y) - 1| < \epsilon$
===== so here is my old inelegant solution to see how convoluted my mind can get...=====
====
What your wrote ($\lvert x^2y\rvert+\lvert y \rvert + 1 < \epsilon$) doesn't work as Soke pointed out, as your expression is greater than 1.
What you want is $|f(x,y) - 1| \le \max(|x^2y - 1|,|y-1|) \le \epsilon$ so ...
$\sqrt{(1 - x)^2 + (1- y)^2} < \delta \implies$
$|1-x| < \delta$ and $|1 - y| < \delta$ so if $\delta \le \epsilon$ we have $|y - 1| < \epsilon$. 
$|1 -x| < \delta \implies |x| < 1 \pm \delta$ and $|1 - y| < \delta \implies |y| < 1 \pm \delta$ so $x^2y < 1 + 3\delta + 3\delta^2 + \delta^3$.
So for any $\epsilon > 0$ chose $\delta$ such that $3\delta + 3\delta^2 + \delta^3 \le \epsilon$.  We don't actually have to solve for $\delta$-- if $\delta < \epsilon/4$ will be good enough for small enough epsilon.
$|(x,y) - (1,1)| < \epsilon/4$
$|y-1| < \epsilon/4 < \epsilon$
$|x^2y - 1| < (1+ \epsilon/4)^3 - 1 = 3\epsilon/4 + 3\epsilon^2/4 + \epsilon^3 <  \epsilon$ for small enough epsilon.
So $|f(x,y) - 1| = \{|y-1|, |x^2y-1|\} < \epsilon$.
</s>