<s>[INST] 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. 
 [/INST] 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 $1.$
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