We are analyzing the given problem, where $f : \mathbb{R}^3 \to \mathbb{R}$, defined as:

$f(x, y, z) = xy^2 - x^2 - y^2,$

and we need to determine the nature of the point $(0, 0, 0)$ based on the given options:

1. $f$ reaches no critical point at $(0,0,0)$,
2. $f$ reaches a relative minimum at $(0,0,0)$,
3. $f$ reaches a relative maximum at $(0,0,0)$,
4. $f$ reaches a saddle point at $(0,0,0)$.

Step 1: Find the critical points

A critical point occurs where the gradient of $f$, $\nabla f(x, y, z)$, is zero. The gradient is given by:

$\nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right).$

Compute partial derivatives:

1. $\frac{\partial f}{\partial x} = y^2 - 2x$
2. $\frac{\partial f}{\partial y} = 2xy - 2y$
3. $\frac{\partial f}{\partial z} = 0 \quad (\text{no } z \text{ dependence in } f).$

Set the gradient to zero: $\nabla f(x, y, z) = (y^2 - 2x, 2xy - 2y, 0) = (0, 0, 0).$

From this system:

1. $y^2 - 2x = 0 \Rightarrow x = \frac{y^2}{2}$,
2. $2y(x - 1) = 0 \Rightarrow y = 0 \text{ or } x = 1$,
3. $z$ is unconstrained ($f$ is independent of $z$).

• If $y = 0$, then $x = 0$.
• If $x = 1$, then $y^2 = 2$.

Thus, the critical points are: $(0, 0, z) \quad \text{and} \quad (1, \pm \sqrt{2}, z).$

Step 2: Classify $(0, 0, z)$

To classify the critical point $(0, 0, z)$, compute the Hessian matrix of $f(x, y, z)$ with respect to $x$ and $y$ (since $z$ does not influence $f$):