To find the stationary point, we need to find where the derivative \(y'\) is zero.

Given \(y = x^{-\frac{2}{3}} \ln x\), we use the product rule to differentiate:

Let \(u = x^{-\frac{2}{3}}\) and \(v = \ln x\).

Then \(u' = -\frac{2}{3}x^{-\frac{5}{3}}\) and \(v' = \frac{1}{x}\).

Using the product rule, \(y' = u'v + uv'\).

So, \(y' = -\frac{2}{3}x^{-\frac{5}{3}} \ln x + x^{-\frac{2}{3}} \cdot \frac{1}{x}\).

Simplifying, \(y' = -\frac{2}{3}x^{-\frac{5}{3}} \ln x + x^{-\frac{5}{3}}\).

Factor out \(x^{-\frac{5}{3}}\):

\(y' = x^{-\frac{5}{3}} \left( -\frac{2}{3} \ln x + 1 \right)\).

Set \(y' = 0\):

\(x^{-\frac{5}{3}} \left( -\frac{2}{3} \ln x + 1 \right) = 0\).

Since \(x^{-\frac{5}{3}} \neq 0\) for \(x > 0\), solve:

\(-\frac{2}{3} \ln x + 1 = 0\).

\(\ln x = \frac{3}{2}\).

\(x = e^{\frac{3}{2}}\).

Substitute \(x = e^{\frac{3}{2}}\) back into the original equation to find \(y\):

\(y = \left( e^{\frac{3}{2}} \right)^{-\frac{2}{3}} \ln \left( e^{\frac{3}{2}} \right)\).

\(y = e^{-1} \cdot \frac{3}{2}\).

\(y = \frac{3}{2e}\).

Thus, the exact coordinates of the stationary point are \(\left( e^{\frac{3}{2}}, \frac{3}{2e} \right)\).