Answer: stationary point \((0.43,0.98)\); area \(2.75\).

Differentiate

\(y=x+6(3x+2)^{-2}.\)

This gives

\(\frac{dy}{dx}=1-36(3x+2)^{-3}.\)

At a stationary point,

\(1-\frac{36}{(3x+2)^3}=0.\)

So

\((3x+2)^3=36.\)

Hence

\(x=\frac{\sqrt[3]{36}-2}{3}=0.4339\ldots.\)

Substitute this into the curve:

\(y=x+\frac{6}{(3x+2)^2}=0.9845\ldots.\)

Therefore the stationary point is

\(\boxed{(0.43,0.98)}.\)

The shaded area is

\(\int_0^2\left(x+\frac{6}{(3x+2)^2}\right)\,dx.\)

An antiderivative is

\(\frac{x^2}{2}-\frac{2}{3x+2}.\)

Therefore the area is

\(\left[\frac{x^2}{2}-\frac{2}{3x+2}\right]_0^2.\)

This equals

\(\left(2-\frac14\right)-\left(0-1\right)=\frac74+1=\frac{11}{4}.\)

So the area is

\(\boxed{2.75}.\)