We start with the equations:

1. \(u = x^2 y\) 2. \(y + 3x = 9\)

Express \(y\) in terms of \(x\):

\(y = 9 - 3x\)

Substitute \(y\) into the equation for \(u\):

\(u = x^2 (9 - 3x) = 9x^2 - 3x^3\)

Differentiate \(u\) with respect to \(x\):

\(\frac{du}{dx} = 18x - 9x^2\)

Set \(\frac{du}{dx} = 0\) to find the stationary points:

\(18x - 9x^2 = 0\)

\(9x(2 - x) = 0\)

\(x = 0\) or \(x = 2\)

Since \(x\) cannot be zero (as \(x\) is non-zero), we have \(x = 2\).

Substitute \(x = 2\) back into \(y = 9 - 3x\):

\(y = 9 - 3(2) = 3\)

Calculate \(u\):

\(u = (2)^2 (3) = 12\)

To determine if this is a maximum or minimum, find the second derivative:

\(\frac{d^2u}{dx^2} = 18 - 18x\)

Evaluate at \(x = 2\):

\(\frac{d^2u}{dx^2} = 18 - 18(2) = -18\)

Since \(\frac{d^2u}{dx^2} < 0\), the stationary point is a maximum.