<s>[INST] Is there a step by step checklist to check if a multivariable limit exists and find its value? Do we rely on certain intuition or is there an unofficial general crude checklist I should follow?

I had a friend telling me that if the sum of the powers on the numerator is smaller then the denominator, there is a higher chance that it may not exists.
And if the sum of powers on the numerator is higher then the denominator, most likely, it exists.
Also if there is a sin or cos or e-constant, it most likely exists.

How can I know what to do at the first glance given so little time exists for me in exam to ponder? If I spend all my time on figuring out a two-path test when the limit exists, that would be a huge disaster.
Is this one of those cases where practice makes perfect?
Example: $$\lim_{(x,y)\to(0,0)}\frac{(\sin^2x)(e^y-1)}{x^2+3y^2}$$
Please give me a hint and where do you get the hint.
Example: $$\lim_{(x,y)\to(0,0)}\exp\left(-\frac{x^2+y^2}{4x^4+y^6}\right)$$
I need a hint for this too.

Common methods I have learnt for reference:
Two-Path test, Polar Coordinates, Spherical Coordinates, Mean Value Theorem using inequalities.

 [/INST] I think there is no common method for all types of limits. You need significantly decrease the range of possible functions to get at least some kind of a road map.
For this two particular limits I suggest you the following two "brand new" approaches:


*

*The first one is usage of equivalences (or more general use of Taylor series expansion). Since $\sin x\sim x$ and $e^y-1\sim y$ we get 
$$
\begin{align}
\lim\limits_{(x,y)\to(0,0)}\frac{\sin^2 x(e^y-1)}{x^2+3y^2}&=
\lim\limits_{(x,y)\to(0,0)}\frac{x^2 y}{x^2+3y^2}\frac{\sin^2 x}{x^2}\frac{e^y-1}{y}\\
&=\left(\lim\limits_{(x,y)\to(0,0)}\frac{x^2 y}{x^2+3y^2}\right)\left(\lim\limits_{(x,y)\to(0,0)}\frac{\sin^2 x}{x^2}\right)\left(\lim\limits_{(x,y)\to(0,0)}\frac{e^y-1}{y}\right)\\
&=\lim\limits_{(x,y)\to(0,0)}\frac{x^2 y}{x^2+3y^2}\\
\end{align}
$$ 
The last limit does exists and equals $0$. Indeed
$$
0\leq \left|\frac{x^2y}{x^2+3y^2}\right|=\frac{x^3|y|}{x^2+3y^2}\leq\frac{x^2|y|}{2\sqrt{3}xy}=\frac{1}{2\sqrt{3}}x\;\mathrm{sign}\; y
$$
$$
\lim\limits_{(x,y)\to(0,0)}\frac{1}{2\sqrt{3}}x\;\mathrm{sign}\; y=0
$$
Hence by mean value theorem
$$
\lim\limits_{(x,y)\to(0,0)}\frac{x^2 y}{x^2+3y^2}=0
$$
so
$$
\lim\limits_{(x,y)\to(0,0)}\frac{\sin^2 x(e^y-1)}{x^2+3y^2}=0
$$

*The second one is usage of continuiuty of fucntions. In your particular case this is $\exp$, then
$$
\lim\limits_{(x,y)\to(0,0)}\exp\left(-\frac{x^2+y^2}{x^4+3y^6}\right)=\exp\left(-\lim\limits_{(x,y)\to(0,0)}\frac{x^2+y^2}{4x^4+y^6}\right)
$$ 
Now we introduce polar coordinates to evaluate the internal limit
$$
\begin{align}
\lim\limits_{(x,y)\to(0,0)}\frac{x^2+y^2}{4x^4+y^6}&=
\lim\limits_{r\to 0}\frac{r^2}{4r^4\cos^4\phi+r^6\sin^6\phi}\\
&=\lim\limits_{r\to 0}\frac{1}{r^2(4\cos^4\phi+r^2\sin^6\phi)}
\end{align}
$$
Again we use inequalities
$$
r^2(4\cos^4\phi+r^2\sin^6\phi)\leq r^2(4\cos^2\phi+r^2\sin^2\phi)\\\leq r^2\max(4,r^2)(\cos^\phi+\sin^2\phi)=r^2\max(4,r^2)
$$
and even more for $r<2$ we can say that
$$
r^2(4\cos^4\phi+r^2\sin^6\phi)\leq r^2\max(4,r^2)\leq 4r^2
$$
hence
$$
\begin{align}
\lim\limits_{(x,y)\to(0,0)}\frac{x^2+y^2}{4x^4+y^6}&=
\lim\limits_{r\to 0}\frac{1}{r^2(4\cos^4\phi+r^2\sin^6\phi)}\\
&\geq\lim\limits_{r\to 0}\frac{1}{4r^2}=+\infty
\end{align}
$$
and
$$
\lim\limits_{(x,y)\to(0,0)}\exp\left(-\frac{x^2+y^2}{x^4+3y^6}\right)=\exp(-\infty)=0
$$
</s>