Answer
$$ \bar{x}=\frac{3}{8} \ \text{ and }\ \bar{y}=\frac{17}{16}$$
Work Step by Step
Since
\begin{align*}
M&=\int_{0}^{1} \int_{x}^{2-x}(6 x+3 y+3) d y d x\\
&=\int_{0}^{1}\left[6 x y+\frac{3}{2} y^{2}+3 y\right]_{x}^{2-x} d x\\
&=\int_{0}^{1}\left(12-12 x^{2}\right) d x\\
&=8
\end{align*}
and
\begin{align*}
M_{y}&=\int_{0}^{1} \int_{x}^{2-x} x(6 x+3 y+3) d y d x\\
&=\int_{0}^{1}\left(12 x-12 x^{3}\right) d x\\
&=3 \\
M_{x}&=\int_{0}^{1} \int_{x}^{2-x} y(6 x+3 y+3) d y d x\\
&=\int_{0}^{1}\left(14-6 x-6 x^{2}-2 x^{3}\right) d x\\
&=\left(14x-3x^2-2 x^{3}-\frac{1}{3} x^{4}\right)\bigg|_{0}^{1} \\
&=\frac{17}{2}
\end{align*}
Then
$$ \bar{x}=\frac{3}{8} \ \text{ and }\ \bar{y}=\frac{17}{16}$$
