Limit with 2 variables: $\frac{x^2y}{x^4 +y^2}$, $\frac{e^{xy^3}-1}{x^2 +y^4}$ and $(x^2 +y^2)^{xy}$ at $(0,0)$ I am new to this topic so I appreciate any help on this.
a)$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{x^2y}{x^4 +y^2}$$
For $x=0 \lor y=0 :  f(x,y) \longrightarrow 0$ but for $y = x^2$ the limit is $\frac{1}{2}$ so the limit does not exist.
b)
$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = \frac{e^{xy^3}-1}{x^2 +y^4}$$
I don't have any ideas here
c)
$$\lim_{(x,y) \rightarrow (0,0)} f(x,y) = (x^2 +y^2)^{xy}$$
Let $x \geq y$ then $|x^2 +y^2|^{xy} \leq |2y^2|^{y^2} \longrightarrow 1 \quad $for $y \longrightarrow 0$ but for $x = \frac{1}{y} \Longrightarrow f(x,y) = (\frac{1}{y^2} + y^2)^1 \longrightarrow \infty \quad$ for $y \longrightarrow 0 $ so the limit does not exist. 
 A: For b):
we have that $f(0,y) = f(x, 0) = 0$ for every $x,y\neq 0$.
If $x$ and $y$ are both non zero, we have that
$$
\frac{e^{xy^3} - 1}{x^2+y^4} = \frac{e^{xy^3} - 1}{xy^3}\cdot \frac{xy^3}{x^2+y^4}.
$$
The first factor at the r.h.s. is bounded (for $x,y$ bounded), since $(e^t-1)/t \to 1$ as $t\to 0$.
The second factor can be estimated by
$$
\left|\frac{xy^3}{x^2+y^4}\right| =
|y| \cdot \frac{|x| y^2}{x^2+y^4}\leq \frac{|y|}{2}.
$$
Hence the limit exists and is equal to $0$.
A: Another way to process is to set $y=mx^\alpha$ with some well chosen $\alpha>0$ so that the various sums $x^a+y^b$ become homogeneous.
Of course $m=m(x,y)$ is not a constant, just the ratio of the rate of convergence of $x$ and $y$.
Part a:
With $y=mx^2$ 
$\displaystyle f(x,y)=\frac{x^2y}{x^4+y^2}=\frac{mx^4}{x^4+m^2x^4}=\frac{m}{1+m^2}$ 
This is bounded but can take multiple values (e.g $m=1,\ f(x,x^2)=\frac 12$ and $m=0,\ f(x,0)=0$), so there is no limit in $(0,0)$.
Part b:
With $x=my^2$ 
$\displaystyle f(x,y)=\frac{e^{x^3y}-1}{x^2+y^4}\sim\frac {x^3y}{x^2+y^4}=\frac{m^3y^7}{m^2y^4+y^4}=\frac{m^3y^3}{1+m^2}=\underbrace{\bigg(\frac{m^2}{1+m^2}\bigg)}_\text{bounded}xy\to 0$
So there is a limit and it is $0$.
Part c:
The previous method do not work well here, but polar coordinates do
$f(x,y)=(x^2+y^2)^{xy}=\exp(xy\ln(x^2+y^2))=\exp(\underbrace{\cos(\theta)\sin(\theta)}_\text{bounded}\underbrace{r^2\ln(r^2)}_{\to 0})\to e^0=1$
A: You've made a mistake in c): $(1/y,y)$ does not approach $(0,0)$ as $y\to 0.$
To do it, we can apply $\ln$ to get $xy\ln (x^2+y^2).$ Recall $|xy|\le (x^2+y^2)/2.$ Thus
$$\tag 1 |xy \ln (x^2+y^2)| \le \frac{(x^2+y^2)\ln (x^2+y^2)}{2}.$$
Now $u\ln u \to 0$ as $u\to 0^+.$ Thus the limit of $(1)$ as $(x,y) \to (0,0)$ is $0.$ This implies the orginal limit is $1.$
