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The problem given is based on the scalar function \( f(x, y) = x^2 - y^3 + x + 3y \). Let's solve it step by step.

### Part (a): Determine the stationary points of \( f(x, y) \).

Stationary points occur when the gradient of the function is zero, i.e., when both partial derivatives with respect to \( x \) and \( y \) are equal to zero.

#### Step 1: Find the partial derivatives.

1. **Partial derivative with respect to \( x \):**

\[
f_x(x, y) = \frac{\partial}{\partial x} (x^2 - y^3 + x + 3y) = 2x + 1
\]

2. **Partial derivative with respect to \( y \):**

\[
f_y(x, y) = \frac{\partial}{\partial y} (x^2 - y^3 + x + 3y) = -3y^2 + 3
\]

#### Step 2: Set both partial derivatives equal to zero to find the stationary points.

- For \( f_x(x, y) = 0 \):

\[
2x + 1 = 0 \implies x = -\frac{1}{2}
\]

- For \( f_y(x, y) = 0 \):

\[
-3y^2 + 3 = 0 \implies y^2 = 1 \implies y = \pm 1
\]

#### Step 3: Find the corresponding \( (x, y) \) points.

From these conditions, the stationary points are:

\[
(x, y) = \left(-\frac{1}{2}, 1\right) \quad \text{and} \quad \left(-\frac{1}{2}, -1\right)
\]

### Part (b): Calculate \( \nabla f \) and the directional derivative of \( f(x, y) \) along the unit normal at the stationary points.

#### Step 1: Compute the gradient \( \nabla f \).

The gradient of \( f(x, y) \) is:

\[
\nabla f(x, y) = \left( f_x(x, y), f_y(x, y) \right) = \left( 2x + 1, -3y^2 + 3 \right)
\]

At the stationary points:

1. For \( \left( -\frac{1}{2}, 1 \right) \):

\[
\nabla f\left( -\frac{1}{2}, 1 \right) = \left( 0, 0 \right)
\]

2. For \( \left( -\frac{1}{2}, -1 \right) \):

\[
\nabla f\left( -\frac{1}{2}, -1 \right) = \left( 0, 0 \right)
\]

Since the gradient at the stationary points is \( \mathbf{0} \), there is no preferred direction for the directional derivative, meaning the directional derivative along any direction (including the normal direction) is also zero.

Thus, the directional derivative along the unit normal is **zero** at both stationary points.

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Would you like more details on the process, or do you have any questions?

Here are five related questions for further exploration:

1. How do you classify stationary points as maxima, minima, or saddle points?
2. What is the significance of the gradient being zero at a stationary point?
3. How can you find the second derivatives to analyze the nature of stationary points?
4. How would the directional derivative behave if the gradient were non-zero?
5. What are the implications of having multiple stationary points in a function?

**Tip:** To classify stationary points, use the second derivative test (Hessian matrix). It helps in determining whether the points are minima, maxima, or saddle points.

Consider the scalar function f(x,y) = x^2 - y^3 + x + 3y. Determine the stationary point(s) of f. Calculate ∇f and the directional derivative of f(x,y) along the unit normal at the stationary points.

This problem explores how to find stationary points of the scalar function f(x,y) = x^2 - y^3 + x + 3y, and compute the directional derivative along the unit normal. It includes solving for partial derivatives and analyzing the gradient at specific points.

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