To find the saddle points for the function $f(x, y) = 9(x^2 + y^2)e^{y^2 - x^2}$, we need to perform the following steps:

1. Find the first-order partial derivatives (the gradient) of the function with respect to $x$ and $y$.
2. Set these derivatives equal to zero to find critical points.
3. Determine the nature of the critical points (whether they are saddle points, maxima, or minima).

Step 1: First-order partial derivatives

The function is:

$f(x, y) = 9(x^2 + y^2) e^{y^2 - x^2}$

Partial derivative with respect to $x$:

We will use the product rule and chain rule to differentiate. The term $9(x^2 + y^2)$ and $e^{y^2 - x^2}$ will both contribute to the derivative.

$f_x = \frac{\partial}{\partial x}\left(9(x^2 + y^2)e^{y^2 - x^2}\right)$

Using the product rule:

$f_x = 9\left( \frac{\partial}{\partial x}(x^2 + y^2)\right)e^{y^2 - x^2} + 9(x^2 + y^2)\frac{\partial}{\partial x}\left(e^{y^2 - x^2}\right)$

Since $\frac{\partial}{\partial x}(x^2 + y^2) = 2x$ and $\frac{\partial}{\partial x}(e^{y^2 - x^2}) = -2xe^{y^2 - x^2}$, we get:

$f_x = 9(2x)e^{y^2 - x^2} + 9(x^2 + y^2)(-2x)e^{y^2 - x^2}$ $f_x = 9e^{y^2 - x^2} \left(2x - 2x(x^2 + y^2)\right)$ $f_x = 18x e^{y^2 - x^2} \left(1 - (x^2 + y^2)\right)$

Partial derivative with respect to $y$:

Similarly, using the product rule for $y$:

$f_y = \frac{\partial}{\partial y}\left(9(x^2 + y^2)e^{y^2 - x^2}\right)$ $f_y = 9\left( \frac{\partial}{\partial y}(x^2 + y^2)\right)e^{y^2 - x^2} + 9(x^2 + y^2)\frac{\partial}{\partial y}\left(e^{y^2 - x^2}\right)$

Since $\frac{\partial}{\partial y}(x^2 + y^2) = 2y$ and $\frac{\partial}{\partial y}(e^{y^2 - x^2}) = 2ye^{y^2 - x^2}$, we get:

$f_y = 9(2y)e^{y^2 - x^2} + 9(x^2 + y^2)(2y)e^{y^2 - x^2}$ $f_y = 18y e^{y^2 - x^2} \left(1 + (x^2 + y^2)\right)$

Step 2: Set the first-order derivatives equal to zero

To find the critical points, we set both $f_x$ and $f_y$ equal to zero.

For $f_x = 0$:

$18x e^{y^2 - x^2} \left(1 - (x^2 + y^2)\right) = 0$ Since $e^{y^2 - x^2} \neq 0$, this simplifies to:

$x \left(1 - (x^2 + y^2)\right) = 0$ This gives two possibilities:

1. $x = 0$
2. $x^2 + y^2 = 1$

For $f_y = 0$:

$18y e^{y^2 - x^2} \left(1 + (x^2 + y^2)\right) = 0$ Since $e^{y^2 - x^2} \neq 0$, this simplifies to:

$y \left(1 + (x^2 + y^2)\right) = 0$ This gives two possibilities:

1. $y = 0$
2. $x^2 + y^2 = -1$, but this has no real solutions.

Step 3: Analyze the critical points

We now analyze the cases where $x = 0$ or $x^2 + y^2 = 1$.

Case 1: $x = 0$

When $x = 0$, the equation $x^2 + y^2 = 1$ gives $y^2 = 1$, so $y = \pm 1$. Thus, the critical points are $(0, 1)$ and $(0, -1)$.

Case 2: $x^2 + y^2 = 1$

For this case, the critical points are the points on the circle defined by $x^2 + y^2 = 1$.

Step 4: Determine the nature of the critical points

We can use the second derivative test to determine whether the critical points correspond to maxima, minima, or saddle points. This involves computing the second-order partial derivatives and evaluating the discriminant $D = f_{xx} f_{yy} - (f_{xy})^2$.

Would you like to proceed with the second derivative test to analyze the nature of these critical points?