Q:

Show that $\lim_{(x,y) \to (0,0)} \frac{x\sqrt{y}}{x+y^2}$ doesn't exist using two-path test

How can I show that the limit:
$$\lim_{(x,y) \to (0,0)} \frac{x\sqrt{y}}{x+y^2}$$
doesn't exist using two-path test?
All the following paths gave a zero:

$f(x,0)$
$f(0,y)$
$f(x,mx)$
$f(x,x^2)$
$f(\sqrt{y},y)$

A:

We want two paths with different outcome, we can easily find that if we take
$$f(x,y)=\frac{x\sqrt{y}}{x+y^{2}},$$
then 
$$\lim_{x\rightarrow0}f(x,0)=0.$$
We now want to find a path where the function is constant $1$, i.e.,
$$f(x,y)=\frac{x\sqrt{y}}{x+y^{2}}=1\Leftrightarrow x=\frac{y^{2}}{\sqrt{y}-1}$$
We define the path $\gamma:[0,1)\rightarrow\mathbb{R}^{2}$ given by
$$\gamma(t)=\left(\frac{t^{2}}{\sqrt{t}-1},t\right).$$
We find
$$\lim_{t\rightarrow 0}f(\gamma(t))=1.$$

