Infinite stationary points?
As I'm only having my first calculus class and not a maths student I cannot answer the question of how to always determine the type of points. However, there can indeed be an infinite of stationary points for a given function.
By letting $f_x(x, y) = 0$ we determine that $y = 0$ must be true. Using this to let $f_y(x, y) = 0$ we find that this is already $0$. In case we don't believe it doesn't matter what $x$ is, let's take an example. If we let $x = \pi$, which has no reason to be chosen. We find that
$$
f_y(\pi, 0) = 2\cdot\pi\cdot0
$$
This is indeed equal to zero, so any point $(x, 0)$, with $x\in\mathbb{R}$, is a stationary point.
Calculating the determinant of the hessian gives that this is zero. So how do we determine the type of stationary point? This can be done by using a bit of intuition, or if possible plotting the function too. If we take a slice of the function where we vary the value of $y$, we can see that the function is of the form $c \cdot y^2$. When $x < 0$, this parabola opens to the bottom, so our point is a maximum. If we look at points where $x > 0$, the parabola opens to the top, so these points are minima. In the special case of $x = 0$, the function is a line. Since it has positive values on one side and negative on the other, we can say that this point is a saddlepoint. We can confirm this by plotting in this case, see
this WolframAlpha calculation*, that what we intuitively thought gives the correct answer. See also the image below:
Plot of function $x \cdot y^2$+
* I used the 'stationary point calculator' for the link, but WolframAlpha does not display all the stationary points we found here. It does plot the function for us.
+ I need 10 reputation to post the image, so I can't post the image, it's just the link now.
A more complex example
This second example is actually a bonus question from my homework this week. I will use a smaller question with sums as an example. Let's first find the stationary point of
$$
h(x_1, \dotsc, x_n) = \sum_{i=1}^n \bigl(x_i - i\bigr)^2
$$
Since this is just the sum of multiple variables, without interaction between them, the partial derivative to any $x_i$ with $1 \leq i \leq n$ is given by the partial derivative to this term:
$$
\frac{\partial}{\partial x_i} \sum_{i=1}^n \bigl(x_i - i\bigr)^2 =
\frac{\partial}{\partial x_i} \Bigl[\bigl(x_i - i\bigr)^2\Bigr] = 2\bigl(x_i - i\bigr)
$$
As for when variables are letters, the stationary point is the point where all partial derivatives are zero. This means that $f_{x_i} = 0$ for all $1 \leq i \leq n$. Solving this gives us the answer that $x_i = i$ which corresponds with the point $(1, 2, 3, \dotsc, n)$
The calculation of the stationary points of the function $g$ can be done very similar. Instead of calculating the partial derivative for each term at the same time, we will look at the cases where $i=1$, $i=2 \vee i=3$ and $i\geq4$. Let us first calculate the partial derivative for $i=1$.
As we can see when we look at the function, the first part is a constant times $x_1$ and the second part contains no $x_1$. This means that our partial derivative is just the constant before $x_1$, the sum of squares.
$$
g_{x_1}(x_1, \dotsc, x_n) = \sum_{i=4}^n x_i^2
$$
The partial derivative for $x_2$ and $x_3$ can be calculated in a similar way. Their derivatives are:
$$
g_{x_2}(x_1, \dotsc, x_n) = g_{x_3}(x_1, \dotsc, x_n) = \sum_{i=4}^n x_i
$$
For $i\geq4$ we find that the partial derivative in a similar fashion to what we did before. Since it is about a sum, we know that we really only need to take the partial derivative for the specific term $i$:
$$
g_{x_i}(x_1, \dotsc, x_n) = \frac{\partial}{\partial x_i} \Bigl[x_1 \cdot x_i^2 + \bigl(x_2+x_3\bigr) x_i\Bigr] = 2 x_1 x_i + x_2 + x_3 \qquad \textrm{where }i\geq4
$$
If we solve $g_{x_1}\! = 0$, we find that all $x_i$ for $i\geq4$ are zero, as there are no reals whose square is negative. Substituting this into $g_{x_2}$ and $g_{x_3}$ yields that these equal to zero now too. This means the last we have to solve is $g_{x_i}\! = 0$.
$$\begin{align}
g_{x_i}(x_1, \dotsc, x_n) &= 0 \\
2 x_1 x_i + x_2 + x_3 &= 0 \\
0 + x_2 + x_3 &= 0 \qquad \textrm{since } x_i = 0 \textrm{ for } i\geq4\\
x_2 &= -x_3
\end{align}$$
This gives us that there are no constraints for $x_1$ and that the only constrains for $x_2$ and $x_3$ are that they should be each other's opposite. The stationary points of $g(x_1, \dotsc, x_n)$ are the points given by $\bigl(a, b, -b, 0, \dotsc, 0\bigr)$ where $a, b \in \mathbb{R}$.
I will not dive into which points are minimum and maximum, though I assume it is going to be similar to what we found with our first example.
