The way I try to prove the uniform continuity of $f(x,y)=x^2+3y$ let $f(x,y)=x^2+3y$ and $0≤x≤1 \\ 0≤y≤1$  Now, To prove uniform continuity given $\epsilon >0$  we need to find a $\delta >0$ such that 
$$|p-q|< \delta \Longrightarrow |f(p)-f(q)| < \epsilon$$
okay on this rectangle in $R^2$ we know that $|p-q|<\sqrt2$ 
Now let $p=(x,y)$ and $q=(x_0,y_0)$
Then $|f(p)-f(q)|=|x^2+3y - x_0^2+3y_0| = |(x-x_0)(x+x_0)+3(y-y_0)|$
since $|x-x_0|≤|p-q|<\sqrt2$
and also $|y-y_0|≤|p-q|≤ \sqrt2$
And $|x-x_0|<1$
We can then conclude that$$ |(x-x_0)(x+x_0)+3(y-y_0)|<|4\sqrt2|< \epsilon$$
Now $\forall \epsilon >0$ choose $\delta = 4 \sqrt 2$ and then we satisfy the definition of uniform continuity.
However, it usually seems like textbooks don't use simple approaches as mine? They usually try to write $\delta$ in terms of $\epsilon$. Is my approach wrong? If my approach isn't wrong, then why textbooks don't use such approaches?
 A: You may want to use the fact that $$|x-x_0| \leq \sqrt{(x-x_0)^2+(y-y_0)^2} = \|p-q\|$$ $$|y-y_0| \leq \sqrt{(x-x_0)^2+(y-y_0)^2}= \|p-q\|$$
Also, remark that in the rectangle you have $|x+x_0|\leq 2$. From the expression you wrote, we get:
$$|f(p)-f(q)| \leq 2\|p-q\| + 3 \|p-q\|= 5\|p-q\|$$
Now just take $\delta = \frac{\epsilon}{5}$
This is if you want to prove it explicitly. Else, you may just use Heine-Cantor theorem.
A: To directly address your question, this does not work. You are trying to show that for every $\epsilon$ given to you, you can find a $\delta$ such that any two points which are less than $\delta$ apart will have a difference of less than $\epsilon$. Your answer as it reads currently suggests that regardless of what $\epsilon$ is given to you, all points within a distance of $4 \sqrt{2}$ from each other will be within the given bound, suggesting that the function is constant, which it is not.
More specifically, your conclusion that $|(x-x_0)(x+x_0)+3(y-y_0)|< 4\sqrt2$ seems correct; however, how does this suggest that $\delta = 4 \sqrt{2}$? 
See @Harnak's answer for an elegant way to do the proof.
