Answer: \(y=5-\ln x-x\); tangent \(y=-\frac43x+6-\ln3\).

(a) The gradient of the normal is

\(\frac{x}{x+1}.\)

The tangent gradient is the negative reciprocal, so

\(\frac{dy}{dx}=-\frac{x+1}{x}.\)

Rewrite this as

\(\frac{dy}{dx}=-1-\frac1x.\)

Integrate:

\(y=-x-\ln x+C.\)

The curve passes through \((1,4)\), so

\(4=-1-\ln1+C.\)

Since \(\ln1=0\),

\(C=5.\)

Therefore

\(y=5-\ln x-x.\)

(b) When \(x=3\),

\(y=5-\ln3-3=2-\ln3.\)

The tangent gradient is

\(-\frac{x+1}{x}=-\frac43.\)

So the tangent is

\(y-(2-\ln3)=-\frac43(x-3).\)

Hence

\(y=-\frac43x+6-\ln3.\)