Answer: \(A=(15,-1)\).

Use differentiation to find the gradient information needed. Stationary points occur when the derivative is zero, and tangents or normals are found from the gradient at the given point.

At \(x=3\),

\(y=\frac{\sqrt{3+1}}{3}=\frac{2}{3}.\)

Differentiate \(y=\frac{\sqrt{x+1}}{x}\):

\(\frac{\mathrm dy}{\mathrm dx}=\frac{x\cdot\frac{1}{2}(x+1)^{-1/2}-\sqrt{x+1}}{x^2}.\)

At \(x=3\), the gradient is

\(\frac{3\cdot\frac{1}{2}(4)^{-1/2}-2}{9}=\frac{\frac34-2}{9}=-\frac{5}{36}.\)

The tangent is therefore

\(y-\frac23=-\frac{5}{36}(x-3)\), so \(y=-\frac{5}{36}x+\frac{13}{12}.\)

At \(A\), this tangent meets \(y=x-16\). Hence

\(x-16=-\frac{5}{36}x+\frac{13}{12}.\)

Multiplying by \(36\) gives \(36x-576=-5x+39\), so \(41x=615\), and \(x=15\).

Then \(y=15-16=-1\). Therefore \(A=(15,-1)\).

This gives the required answer: \(A=(15,-1)\).