I was given this exercise in a exam and I'd really like to know if I managed to accomplish something. I started separating the limit into two summands:
$$\lim_{(x,y)\to (0,0)} x^4(y-1)\ln\bigg(1+\frac{1}{x^2+y^2}\bigg) \ \ \ \ \ + \ \ \ \ \ \ \ a \ \ \lim_{(x,y) \to (0,0)} \frac{\sin(x)y^{3/2}}{(\sqrt{x^2+y^2})^3}$$
It was easy to prove that the first limit equals zero.
$$\bigg| x^4(y-1)\ln\bigg(1+\frac{1}{x^2+y^2}\bigg) \bigg| \leq |y-1|\bigg| x^4 + \frac{x^4}{\|(x,y)\|^2}\bigg| \stackrel{\delta<1}{\leq} |2x^2|\leq|2x|<2\delta$$
But when I tried to bound the second one, I got:
$$\frac{|\sin(x)y^{3/2}|}{\|(x,y)\|^3} \leq \frac{|x||y|^{3/2}}{\| (x,y) \|^3} \leq \frac{\|(x,y)\|^{5/2}}{\|(x,y)\|^3} = \frac{1}{\|(x,y)\|}$$
Which is unbounded when $(x,y) \to (0,0)$.
So... $a$ should be 0?
I should've just shown that approaching (0,0) through $\alpha(t)=(t,t)$ the second limit doesn't exist:
$$\lim_{t\to0^+} \frac{\sin(t)}{t} \cdot \frac{t^{5/2}}{\sqrt{2}t^3} = \lim_{t\to0^+} \frac{t^{5/2}}{\sqrt{2}t^3} = \lim_{t\to0^+} \frac{1}{\sqrt{2t}} = \nexists$$