Unique critical point does not imply global maximum/global minimum

Your intuitive picture is probably as follows: Imaging filling water into the hollow at the origin. As the water rises, since there are points lower than the origin, you expect the water to start spilling over into the lower lying areas at some point. And that point would need to be a critical point! The way to resolve the paradox is to note that before any spilling over occurs, the lake will in fact extend off to infinity. Look at a cross section of your function graph for fixed $y$, as $y$ starts at $0$ and then grows. In other words, study the function $x\mapsto x^2+y^2(1-x)^3$ as $y^2$ grows. You will see a local minimum, starting at the origin and approaching $x=1$ from below, and a local maximum, starting at infinity and approaching $x=1$ from above. As $y^2$ grows, the value at the local minimum grows, while the value at the local maximum decreases. So the aforementioned lake will extend out to $(1,\pm\infty)$ and then start to spill over the edge (the local maximum) at infinity before it spills over anywhere else.

A picture is tremendously helpful. If you have access to maple, try this:

plots[animate](plot,
  [x^2+y^2*(1-x)^3, x = 0 .. 2, view = [0 .. 2, -1 .. 2]], y = 0 .. 10);

Or you can take a look at this static plot of $f(x,y)$. It is easy to see how $(0,0)$ is a local, but not a global, maximum:

I know that f has a minimum when restricted to a compact set. Then (0,0) is a global minimum in any closed ball around (0,0)

This is where your reasoning fails. A global minimum must be attained on a critical point... or at the boundary the set! A much simpler and clearer example could be the function $f(x) = x^3$. It has only one critical point, 0, but when you take a closed ball containing 0, the minimum and maximum that $f$ attains is not at a critical point but at a boundary of the ball.

There is a nice result showing that it is the case for polinomials of low degree.

If f(x,y) is a polynomial of degree $\leq 4$ having a local extrema as the unique critical point, then f(x,y) has a global extrema at that point.

This was proven in Calvert and Vamanamury "Local and global extrema for functions of several variables" J. Austral. Math. Soc. 29, 362-368 (1980)

Consider the following one dimensional example $$ f: \Bbb{R} \to \Bbb{R},\ f(x)=\begin{cases} x^2 & x\geq 0 \\ -x^2 & x <0 \end{cases}$$

This function is differentiable, and it is easy to see that the unique critical point is $x_0=0$. But $x_0=0$ is not a global (not even local) minimum/maximum.

In general, finding critical points is not enough to find the local extrema. You need to use the second derivative test to see whether the critical points are local extrema points.