Answer: \(\left(\frac25,4\right)\).

A stationary point occurs when

\(\frac{\mathrm{d}y}{\mathrm{d}x}=0.\)

So

\((5x-2)^{\frac13}=0,\)

which gives

\(x=\frac25.\)

To find the corresponding \(y\)-coordinate, integrate:

\(y=\int (5x-2)^{\frac13}\,\mathrm{d}x.\)

Using the reverse chain rule,

\(y=\frac{3}{20}(5x-2)^{\frac43}+c.\)

The curve passes through \(\left(2,\frac{32}{5}\right)\), so

\(\frac{32}{5}=\frac{3}{20}(8)^{\frac43}+c.\)

Since \(8^{\frac43}=16\),

\(\frac{32}{5}=\frac{12}{5}+c,\)

so \(c=4\).

At \(x=\frac25\),

\(y=\frac{3}{20}(0)^{\frac43}+4=4.\)

Therefore the stationary point is

\(\left(\frac25,4\right).\)