Answer
$$ \bar{x}=0,\ \ \bar{y}=\frac{32}{45} $$$$ I_{0}=I_{x}+I_{y}=\frac{6}{5}$$
Work Step by Step
Since
\begin{align*}
M&=\int_{0}^{1} \int_{-y}^{y}\left(3 x^{2}+1\right) d x d y\\
&=\int_{0}^{1}\left(2 y^{3}+2 y\right) d y\\
&=\left(\frac{1}{2} y^{4}+ y^2\right)\bigg|_{0}^{1}\\
&=\frac{3}{2} \\
M_{x}&=\int_{0}^{1} \int_{-y}^{y} y\left(3 x^{2}+1\right) d x d y\\
&=\int_{0}^{1}\left(2 y^{4}+2 y^{2}\right) d y\\
&= \left(\frac{2}{5} y^{5}+\frac{2}{3} y^{3}\right)\bigg|_{0}^{1}\\
&=\frac{16}{15}\\
M_{y}&=\int_{0}^{1} \int_{-y}^{y} x\left(3 x^{2}+1\right) d x d y=0
\end{align*}
Then
$$ \bar{x}=0,\ \ \bar{y}=\frac{32}{45} $$
\begin{align*}
I_{x}&=\int_{0}^{1} \int_{-y}^{y} y^{2}\left(3 x^{2}+1\right) d x d y\\
&=\int_{0}^{1}\left(2 y^{5}+2 y^{3}\right) d y\\
&=\left(\frac{1}{3} y^{6}+\frac{1}{2} y^{3}\right)\bigg|_{0}^{1}\\
&=\frac{5}{6}\\
I_{y}&=\int_{0}^{1} \int_{-y}^{y} x^{2}\left(3 x^{2}+1\right) d x d y\\
&=2 \int_{0}^{1}\left(\frac{3}{5} y^{5}+\frac{1}{3} y^{3}\right) d y\\
&=2\left(\frac{3}{30} y^{6}+\frac{1}{12} y^{4}\right)\bigg|_{0}^{1}\\
&=\frac{11}{30}
\end{align*}
Then
$$ I_{0}=I_{x}+I_{y}=\frac{6}{5}$$
