$\require{cancel}$
Find for which values of $\alpha\in\mathbb{R^+}$ the following function is continuous in $\mathbb{R^2}$.
$$f(x,y)=\begin{cases} \frac{x^2+y^2}{|x|^{\alpha}y}\sin\left(\frac{|x|^{\alpha}y}{x^2+y^2}\right), & \text{if $x\neq0$ and $y\neq0$} \\[2ex] 1, & \text{if $x=0$ or $y=0$} \end{cases}$$
I first started by evaluating the limit, $$ \lim_{(x,y)\to (0,0)}\frac{|x|^{\alpha}y}{x^2+y^2}\to \textstyle \text{Polar Substitution} \to \displaystyle \lim_{r\to 0}\frac{|r|^{\alpha}\cdot\left|\cos(\theta)\right|^{\alpha}\cdot \cancel{r}\sin(\theta)}{r^\cancel{2}}= \\ = \displaystyle \lim_{r\to 0}\left(|r|^{\alpha-1}\cdot\left|\cos(\theta)\right|^{\alpha}\sin(\theta)\right)=\begin{cases} 0, & \text{if $\alpha>1$} \\[2ex] \text{Indeterminate}, & \text{if $\alpha\leq1$} \end{cases}$$ And used this result to compute the limit, $$ \lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{|x|^{\alpha}y}\sin\left(\frac{|x|^{\alpha}y}{x^2+y^2}\right)\to \textstyle \begin{align} &u=\frac{|x|^{\alpha}y}{x^2+y^2} \\ &u\to0\text{ as }(x,y)\to(0,0) \end{align} \to \displaystyle \lim_{u\to 0}\frac{\sin(u)}{u}=1$$ which can only be done if $\alpha>1$.
Therefore $f(x,y)$ is continuous $\{\forall\alpha\in\mathbb{R^+}\mid \alpha>1\}$.
Yes, it is correct. However you may express finding the limit $\sin(u)/u$ for $u\to0$ and reducing the question to checking for which $\alpha$ your $u$ tends to $0$.