Find the critical points of the function f(x,y) = 1 − cuberoot(x2 + y2).

The given function is $f(x,y) = 1 - \sqrt[3]{x^2 + y^2}$.
We can rewrite this as $f(x,y) = 1 - (x^2 + y^2)^{1/3}$.

Differentiate $f(x,y)$ with respect to $x$ (treating $y$ as a constant):

$$f_x = \frac{\partial}{\partial x} \left(1 - (x^2 + y^2)^{1/3}\right)$$ $$f_x = 0 - \frac{1}{3}(x^2 + y^2)^{(1/3)-1} \cdot \frac{\partial}{\partial x}(x^2 + y^2)$$ $$f_x = -\frac{1}{3}(x^2 + y^2)^{-2/3} \cdot (2x)$$ $$f_x = -\frac{2x}{3(x^2 + y^2)^{2/3}}$$

Differentiate $f(x,y)$ with respect to $y$ (treating $x$ as a constant):

$$f_y = \frac{\partial}{\partial y} \left(1 - (x^2 + y^2)^{1/3}\right)$$ $$f_y = 0 - \frac{1}{3}(x^2 + y^2)^{(1/3)-1} \cdot \frac{\partial}{\partial y}(x^2 + y^2)$$ $$f_y = -\frac{1}{3}(x^2 + y^2)^{-2/3} \cdot (2y)$$ $$f_y = -\frac{2y}{3(x^2 + y^2)^{2/3}}$$

Critical points occur where $f_x = 0$ and $f_y = 0$, or where $f_x$ or $f_y$ are undefined.

If we set $f_x = 0$:

$$-\frac{2x}{3(x^2 + y^2)^{2/3}} = 0$$

This implies $2x = 0$, so $x = 0$.

If we set $f_y = 0$:

$$-\frac{2y}{3(x^2 + y^2)^{2/3}} = 0$$

This implies $2y = 0$, so $y = 0$.

However, if $x=0$ and $y=0$, the denominators $3(x^2 + y^2)^{2/3}$ become $3(0^2 + 0^2)^{2/3} = 0$. Division by zero means that $f_x$ and $f_y$ are undefined at $(0,0)$.
Therefore, $(0,0)$ is the only critical point.

The second derivative test (D-test) cannot be used here because the second partial derivatives are also undefined at $(0,0)$ due to the $(x^2+y^2)^{-5/3}$ term in their denominators.
Instead, we analyze the behavior of the function directly around $(0,0)$.

The function is $f(x,y) = 1 - \sqrt[3]{x^2 + y^2}$.
Let $r = \sqrt{x^2+y^2}$. This represents the distance from the origin $(0,0)$ to the point $(x,y)$.
Since $r \ge 0$, we have $r^2 = x^2+y^2 \ge 0$.
The term $\sqrt[3]{x^2+y^2} = (x^2+y^2)^{1/3}$ is always non-negative.
The minimum value of $(x^2+y^2)^{1/3}$ occurs when $x^2+y^2=0$, which is at $(0,0)$.
At $(0,0)$, $(0^2+0^2)^{1/3} = 0$.
So, $f(0,0) = 1 - 0 = 1$.

For any other point $(x,y) \ne (0,0)$, we have $x^2+y^2 > 0$.
This means $(x^2+y^2)^{1/3} > 0$.
Therefore, for any $(x,y) \ne (0,0)$, $f(x,y) = 1 - (x^2+y^2)^{1/3} < 1$.
Since $f(x,y) < f(0,0)$ for all $(x,y) \ne (0,0)$, the critical point $(0,0)$ is a local maximum.

The value of the function at this local maximum is $f(0,0) = 1$.

Summary of results:

• Local minima: There are no local minima.
• Local maxima: There is a local maximum at $\boxed{(0,0)}$ with a value of $f(0,0) = \boxed{1}$.
• Saddle points: There are no saddle points.