I have to study the $L^2(\Omega)$ integrability ($\Omega\subset\mathbb{R}^2$ bounded) of the function: $$ f(x,y)= \begin{cases} \dfrac{1}{(1+y^2)|x|^{1/3}} & x\neq 0\\ \\ 0 & x=0 \end{cases} $$ My attempt: Consider the following integral $$ \iint_{\Omega}\bigg(\dfrac{1}{(1+y^2)|x|^{1/3}}\bigg)^2dxdy $$
we have that:
$$ \bigg|\bigg(\dfrac{1}{(1+y^2)|x|^{1/3}}\bigg)^2\bigg|\leq\dfrac{1}{|x|^{2/3}} $$
Now $\Omega$ (being bounded) is contained in some square $[-R,R]\times[-R,R]$ and hence:
$$ \begin{split} \iint_{\Omega}\bigg(\dfrac{1}{(1+y^2)|x|^{1/3}}\bigg)^2dxdy&\leq\iint_{\Omega}\dfrac{1}{|x|^{2/3}}dxdy\\ &\leq\int_{-R}^R\int_{-R}^R\dfrac{1}{|x|^{2/3}}dxdy\\ &=2R\int_{-R}^R\dfrac{1}{|x|^{2/3}}dx<\infty \end{split} $$ This proves that $f\in L^2(\Omega)$.
EDIT: I have also to check if $\partial_x f\in\mathrm{L}^2(\Omega)$; first of all we have that: $$ \partial_x f=-\dfrac{\text{sgn(x)}}{3(1+y^2)|x|^{4/3}} $$
Now if $\overline{\Omega}$ does not contain points with $x=0$ the function $\partial_x f\in\mathrm{L}^2(\Omega)$ being $|\partial_x f|^2$ continuous in $\Omega$ bounded. At this point I don't know how to proceed in the case on which points with $x=0$ are in the domain $\overline{\Omega}$. My idea is to consider $R$ small s.t. $[-R,R]\times[-R,R]\subset\overline{\Omega}$ and then: $$ \iint_\Omega|\partial_x f|^2dxdy\geq\int_{-R}^R\int_{-R}^R\dfrac{1}{9(1+y^2)^2|x|^{8/3}}dxdy =\infty $$ since the integral of $\frac{1}{|x|^{8/3}}$ diverges when $x=0$.
Is it fine?