Answer: \(P=\left(\frac{4+3\ln4}{7},\frac{4+3\ln4}{7}\right)\).

We start with the main method. Differentiate to obtain the tangent gradient, then use the negative reciprocal for the normal gradient where needed.

The curve is

\(y=\ln(x^3+3).\)

Differentiate:

\(\frac{\mathrm dy}{\mathrm dx}=\frac{3x^2}{x^3+3}.\)

At \(x=1\),

\(\frac{\mathrm dy}{\mathrm dx}=\frac34,\)

and

\(y=\ln(1^3+3)=\ln4.\)

The normal gradient is the negative reciprocal of \(\frac34\), so it is

\(-\frac43.\)

The equation of the normal is

\(y-\ln4=-\frac43(x-1).\)

The normal meets \(y=x\) at \(P\), so substitute \(y=x\):

\(x-\ln4=-\frac43(x-1).\)

Multiply by \(3\):

\(3x-3\ln4=-4x+4.\)

Hence

\(7x=4+3\ln4,\)

so

\(x=\frac{4+3\ln4}{7}.\)

Since \(P\) lies on \(y=x\),

\(P=\left(\frac{4+3\ln4}{7},\frac{4+3\ln4}{7}\right).\)

This completes the solution and gives the required result.