Answer
$$\left(\bar{x}, \bar{y}\right) =\left(\frac{\sqrt{10}-1}{\ln (3+\sqrt{10})} ,\frac{\tan ^{-1}(3)}{2 \ln (3+\sqrt{10})}\right)$$
Work Step by Step
Given $$f(x)=\left(1+x^{2}\right)^{-1 / 2}, \quad[0,3]$$
Since
\begin{aligned}
M_{y} &=\rho \int_{a}^{b} x f(x) d x \\
&=\int_{0}^{3} x \frac{1}{\sqrt{1+x^{2}}} d x\\
&=\sqrt{1+x^2}\bigg|_{0}^{3}\\
&=\sqrt{10}-1
\end{aligned}
and
\begin{aligned}
M_{x} &=\frac{1}{2} \rho \int_{a}^{b} f(x)^{2} d x \\
&=\frac{1}{2} \int_{0}^{3}\left(\frac{1}{\sqrt{1+x^{2}}}\right)^{2} d x \\
&=\frac{1}{2} \int_{0}^{3} \frac{1}{1+x^{2}} d x \\
&=\left.\frac{1}{2}\left(\tan ^{-1}(x)\right)\right|_{0} ^{3} \\
&=\frac{1}{2} \tan ^{-1}(3)
\end{aligned}
and
\begin{align*}
M&=\rho A\\
&=\int_{0}^{3} \frac{1}{\sqrt{1+x^{2}}} d x\\
&=\sinh^{-1}(x)\bigg|_{0}^{3}\\
&= \sinh^{-1}(3)\\
&=\ln (\sqrt{10} +3)
\end{align*}
then
\begin{align*}
\left(\bar{x}, \bar{y}\right)&=\left(\frac{M_{y}}{M},\frac{M_{x}}{M} \right)\\
&=\left(\frac{\sqrt{10}-1}{\ln (3+\sqrt{10})} ,\frac{\tan ^{-1}(3)}{2 \ln (3+\sqrt{10})}\right)
\end{align*}
