Answer: \(P=\left(0,\frac{73}{39}\right)\).

Differentiate using the product rule:

\(y=(x-2)(3x+1)^{2/3}.\)

So

\(\frac{dy}{dx} =(3x+1)^{2/3}+(x-2)\cdot\frac23(3x+1)^{-1/3}\cdot3.\)

That is,

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

When \(x=\frac73\), \(3x+1=8\), so

\(y=\left(\frac73-2\right)8^{2/3}=\frac13\cdot4=\frac43.\)

Also,

\(\frac{dy}{dx}=8^{2/3}+2\left(\frac13\right)8^{-1/3} =4+\frac{2}{3}\cdot\frac12 =4+\frac13 =\frac{13}{3}.\)

The gradient of the normal is the negative reciprocal:

\(-\frac{3}{13}.\)

Using the point \(\left(\frac73,\frac43\right)\), the normal is

\(y-\frac43=-\frac{3}{13}\left(x-\frac73\right).\)

At the \(y\)-axis, \(x=0\). Therefore

\(y-\frac43=-\frac{3}{13}\left(-\frac73\right)=\frac7{13}.\)

So

\(y=\frac43+\frac7{13}=\frac{52}{39}+\frac{21}{39}=\frac{73}{39}.\)

Hence

\(\boxed{P=\left(0,\frac{73}{39}\right)}.\)