Find the maxima, minima, and saddle points of the function f(x,y) = 1/(x2+y2-1).

To find the maxima, minima, and saddle points of the function $f(x,y) = \frac{1}{x^2+y^2-1}$, we follow these steps:

Step 1: Find the first partial derivatives and set them to zero to find critical points.
The given function is $f(x,y) = (x^2+y^2-1)^{-1}$.
The partial derivative with respect to $x$ is:

$$f_x = \frac{\partial}{\partial x}(x^2+y^2-1)^{-1} = -1(x^2+y^2-1)^{-2}(2x) = \frac{-2x}{(x^2+y^2-1)^2}$$

The partial derivative with respect to $y$ is:

$$f_y = \frac{\partial}{\partial y}(x^2+y^2-1)^{-1} = -1(x^2+y^2-1)^{-2}(2y) = \frac{-2y}{(x^2+y^2-1)^2}$$

Set $f_x = 0$ and $f_y = 0$:

1. $\frac{-2x}{(x^2+y^2-1)^2} = 0 \implies -2x = 0 \implies x = 0$
2. $\frac{-2y}{(x^2+y^2-1)^2} = 0 \implies -2y = 0 \implies y = 0$ The only critical point is $(0,0)$. Note that the function is undefined when $x^2+y^2-1=0$, i.e., $x^2+y^2=1$. The critical point $(0,0)$ is not on this undefined region.

Step 2: Find the second partial derivatives.

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