The question is to find, for the function $f$ defined below, i) $\alpha \in \mathbb{R}$ such that $f$ is continuous at $(0,0);$ ii) $\alpha \in \mathbb{R}$ such that $f$ is differentiable at $(0,0).$
$$ f(x,y) := \left\{ \begin{array}{rl} \frac{|x|^\alpha |y|^{1/2}}{|x|^{3/2}+y^2}\qquad &\mbox{ if $(x,y)\neq(0,0)$} \\ 0 &\mbox{if $ (x,y)=(0,0)$} \end{array} \right. $$
WLOG $(x,y)\to (0,0)$ within the closed first quadrant. There is a nice result that says for $a,b,c,d > 0$
$$\frac{x^ay^b}{x^c + y^b} \to 0 \iff \frac{a}{c}+\frac{b}{d} > 1.$$
In this problem this amounts to $\alpha > 9/8.$ Thus $f$ is continuous at $(0,0)$ iff $\alpha > 9/8.$
For differentiability, note $f$ is $0$ on the axes, so both partial derivatives of $f$ at the origin are $0.$ Thus if $Df(0,0)$ exists, it must be the zero linear transformation. This amounts to showing
$$\tag 1\frac{x^\alpha y^{1/2}}{(x^{3/2} + y^2)(x^2+y^2)^{1/2}} \to 0.$$
We're fine along the axes. Off the axes let's use polar coordinates, writing $c = \cos t, s = \sin t, ) < t < \pi/2.$ We get
$$\tag 2 r^{\alpha-2}\frac{c^\alpha s^{1/2} }{c^{3/2} + r^{1/2}s^2} \le r^{\alpha-2}\frac{c^\alpha s^{1/2} }{c^{3/2}} = r^{\alpha-2}c^{\alpha-2} (cs)^{1/2}.$$
If $\alpha > 2,$ $(2)\to 0.$ If $\alpha = 2,$ take $x=y$ in $(1)$ to see the limit is not zero. Worse things happen if $\alpha < 2.$ Thus $Df(0,0)$ exists iff $\alpha > 2.$ That's not a result I was expecting; I would have guessed the range would be $\alpha >9/8 + 1 = 17/8.$
