Multivariable delta-epsilon limit proofs The Question asks: Find the delta epsilon limit if it exist
$$ \lim \limits_{(x, y) \to (0, 0)} \ \frac{x^2 y^3}{2x^2 + y^2} = 1$$
What I've done so far:
$$ \mid {\frac{x^2 y^3}{2x^2 + y^2} - 1 \mid} < \epsilon \quad \quad \quad \quad0 < \sqrt{x^2 + y^2} < \delta $$
Now, since $x ^2 \leq 2x^2 + y^ \implies \frac{x^2}{2x^2 + y^2} \leq 1$
Then:
$$ \frac{x^2 \mid y^3 \mid}{2x^2 + y^2} \leq \mid y^3 \mid = \sqrt{y^2}^3 = \quad ...$$
Here is where I am stuck, I can't make $\sqrt{y^2}^3$ into $\sqrt{ x^2 + y^2} $. 
How should I proceed further? 
Using Stewart's Calculus (8th ed) 14.2 Limit proof method
 A: For both $x,y\ne 0$, then
\begin{align*}
\dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}&\leq\dfrac{|x|^{2}|y|^{3}}{2\sqrt{2}|x||y|}\\
&=\dfrac{1}{2\sqrt{2}}|x||y|^{2}\\
&\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2}),
\end{align*}
if either $x=0$ or $y=0$ (but not both), the above inequality is still true. So 
\begin{align*}
\dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2}),~~~~(x,y)\ne(0,0).
\end{align*}
Given $\epsilon>0$, for all $0<\sqrt{|x|^{2}+|y|^{2}}<2^{1/2}\epsilon^{1/3}$, then $\dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2})=\dfrac{1}{2\sqrt{2}}(|x|^{2}+|y|^{2})^{3/2}<\epsilon$. 
The limit is zero.
Another way:
\begin{align*}
\dfrac{x^{2}}{2x^{2}+y^{2}}\leq 1,~~~~(x,y)\ne(0,0),
\end{align*}
so
\begin{align*}
\dfrac{x^{2}|y|^{3}}{2x^{2}+y^{2}}\leq|y|^{3}=(y^{2})^{3/2}\leq(x^{2}+y^{2})^{3/2}.
\end{align*}
A: In fact
$$\left|\frac{x^2y^3}{2x^2+y^2}\right| \leq \frac{1}{2}|y^3|$$ which is equivalent to
$$2x^2\leq 2x^2+y^2.$$ 
