How do I rigorously show that $f(x, y) = \frac{x}{2|x|\sqrt{|x|+|y|}}$ is continuous when $x, y \neq 0$? For the function $f : \mathbb{R}^2 \to \mathbb{R}$ to be continuous, I need to show that for some given $\epsilon > 0$, there exists a $\delta > 0$ so that if $||z - z'|| < \delta$, then $|f(z) - f(z')| < \epsilon$.
Here's what I tried. Let $z = (x, y)$ and $z' = (x', y')$ where $||z - z'|| < \delta$. Then (sorry the formatting is a bit cramped)
\begin{align}
|f(z) - f(z')|
&= \left| \frac{x}{2|x|\sqrt{|x|+|y|}} - \frac{x'}{2|x'|\sqrt{|x'|+|y'|}}\right| \\\\
&= \left| \frac{x |x'|\sqrt{|x'|+|y'|} - x' |x|\sqrt{|x|+|y|}}{2|x|\sqrt{|x|+|y|} |x'|\sqrt{|x'|+|y'|}}\right| \\\\
&\leq \left| x |x'|\sqrt{|x'|+|y'|} - x' |x|\sqrt{|x|+|y|} \right| \\\\
&\leq \left| x |x'|\sqrt{|x'|+|y'|} \right| + \left|x' |x|\sqrt{|x|+|y|} \right| \\\\
&= |xx'| \left| \sqrt{|x'|+|y'|} + \sqrt{|x|+|y|}\right|
\end{align}
but then I get stuck. I know that 
\begin{align}
||z - z'||
&= \sqrt{(x - x')^2 + (y - y')^2} \\
&\leq |x - x'| + |y - y'| \\
&\leq |x| + |x'| + |y| + |y'|
\end{align}
but I'm not sure how to use that to find some $\delta$ that only depends on $\epsilon, x$, and $x'$. 
 A: Theorems are made for that.
In order to show that your $f$ is continuous the simplest and more rigorous proof is to show:
Lemma 1) composition of continuous functions is continuous
Lemma 2) product of continuous functions is continuous
Lemma 3) sum of continuous functions is continuous
Lemma 4) constant, $\sqrt x$ and $|x|$ are continuous functions
Lemma 5) the reciproque of a never vanishing continuous function is continuous
Then the continuity of $f$ easily follows from these lemmata and each one is much easier to prove than the continuity of your original $f$.
A: User126154 has succinctly and eloquently provided the mathematically mature way forward.  
Here, we will illustrate (1) two points at which the proof in the posted question went astray, and (2) one way forward using a "brute force," $\epsilon-\delta$ approach.

In the original posted question, the first point at which the proof went astray occurs at the inequality 
$$
\left| \frac{x |x'|\sqrt{|x'|+|y'|} - x' |x|\sqrt{|x|+|y|}}{2|x|\sqrt{|x|+|y|} |x'|\sqrt{|x'|+|y'|}}\right| 
\leq \left| x |x'|\sqrt{|x'|+|y'|} - x' |x|\sqrt{|x|+|y|} \right|
$$
In general, this inequality $2|x|\sqrt{|x|+|y|} |x'|\sqrt{|x'|+|y'|}\ge 1$  need not hold.  In fact, this term can be arbitrarily small.
The second point at which the proof went astray occurs at the application of the triangle inequality.  While the statement is correct, application of the triangle inequality provides a gross overestimate.  This overestimate does not provide a viable way forward toward proving continuity.

So, how does one proceed in a "brute force" way.  We will look at a simpler problem that will provide the prototype to handle the  problem of interest.
Let's show that $1/\sqrt{x}$ is continuous for $x> 0$.  To that end, we write
$$\begin{align}
\left|1/\sqrt{x}-1/\sqrt{x_0}\right|&=\left|\frac{\sqrt{x_0}-\sqrt{x}}{\sqrt{x}\sqrt{x_0}}\right|\\\\
&=\left|\frac{x-x_0}{x_0\sqrt{x}+x\sqrt{x_0}}\right|
\end{align}$$
Here, we enforce the following.  We will restrict $0<|x-x_0|<a x_0$, where $0<a<1$.  For example, let's choose $a=1/2$ (the choice is arbitrary).  Then, note that 
$$\frac{1}{x_0\sqrt{x}+x\sqrt{x_0}}<1/x_0^{3/2}$$
Using this inequality reveals that for fixed $x_0>0$, given $\epsilon >0$
$$\begin{align}
\left|1/\sqrt{x}-1/\sqrt{x_0}\right|&\le \left|\frac{x-x_0}{x_0^{3/2}}\right|\\\\
&< \epsilon
\end{align}$$
whenever $|x-x_0|< x_0^{3/2}\epsilon$.  Now, let us choose $\delta =\min(x_0^{3/2}\epsilon , x_0/2)$.  Then, for any fixed $x_0>0$, for all $\epsilon >0$, we have 
$$\left|1/\sqrt{x}-1/\sqrt{x_0}\right| < \epsilon$$
whenever $|x-x_0|<\delta =\min(x_0^{3/2}\epsilon , x_0/2)$
This same approach can be used to the problem of interest.
A: Define $L_m$ by {(x, mx); m real} and take all the sequences ($x_n$,$y_n$) converging to ($x_0$,$y_0$) each of them following the paralelle route of some $L_m$. Then f(($x_n$,$y_n$)) converges to f(($x_0$,$y_0$)) for all of these sequences . It is enough. 
