Critical Point of $\mathbb{R}^2 \to \mathbb{R}^2$ function Given a function $f:\mathbb{R}^2 \to \mathbb{R}$ I can find critical points by finding the $1\times 2$ Jacobian matrix, setting each partial derivative equal to zero and solving the equations. I can then classify them by finding the $2\times 2$ Hessian matrix and calculating the eigenvalues. 
But now suppose I have a function $f:\mathbb{R}^2 \to \mathbb{R}^2$, say, $f(x, y) = (y \cos(x), y \sin(x))$. How would I then find and classify critical points? 
Which partial derivatives do I calculate and set to zero? Would my Jacobian matrix now be a $2\times 2$ matrix? And if so then what would the Hessian matrix be like?
 A: If you have $f: \Bbb R^n \to \Bbb R^k$ a differentiable function, then the differential at $p$ is the linear map ${\rm d}f_p: \Bbb R^n \to \Bbb R^k$ such that $$f(p+h)=f(p)+{\rm d}f_p(h) + r(h),$$ with $r(h)/\|h\| \to 0$ as $h \to 0$. We say that $p$ is a critical point of $f$ if ${\rm d}f_p$ is not surjective. Since ${\rm d}f_p$ is a linear map, we can consider its matrix in the canonical basis of $\Bbb R^n$ and $\Bbb R^k$. It will be a $k \times n$ matrix - the Jacobian matrix. You take all the partial derivatives, set them to zero, and solve. $$Jf(p) = [{\rm d}f_p]_{\mathsf{can},\mathsf{can}} = \begin{bmatrix} \frac{\partial f_i}{\partial x_j}\end{bmatrix}_{1 \leq i \leq k, \, 1 \leq j \leq n}$$
Note also that for $k = 1$, we have $Jf(p) = \nabla f(p)$.
The Hessian, as far as I know, it is only good for maps $\Bbb R^n \to \Bbb R$. In $\Bbb R^n$, $n > 1$, we don't have an order relation $\leq$, so we can't talk about "minimum" or "maximum" points. 
However, you can consider the map $$\begin{align}{\rm d}f:& \Bbb R^n \to L(\Bbb R^n, \Bbb R^k) \\ p &\mapsto {\rm d}f_p \in L(\Bbb R^n, \Bbb R^k) \end{align}$$ and ask yourself if this one is differentiable itself. Then you can start thinking of second derivatives and such, and you can identify ${\rm d}^2f_p$ with a bilinear form $B(\cdot, p)$, or something like this, which I think that would be the closest you get to a Hessian.
But I digress. In your example $f: \Bbb R^2 \to \Bbb R^2$ is given by $(y\cos x, y \sin x)$. So $$Jf(x,y) = \begin{bmatrix}-y \sin x & \cos x \\ y \cos x & \sin x  \end{bmatrix}$$ Since $$2 = \dim \ker {\rm d}f_p + \dim {\rm Im}\,{\rm d}f_p,$$ our chances are that $\dim {\rm Im}\,{\rm d}f_p = 0$ or $1$. It can't be zero because $\cos x$ and $\sin x$ can't be zero at the same time. Every point $(x,0)$ will be a critical point because the first column will be zero. Then you would have to check if there are values $(x,y)$ such that the two columns are multiples - this does not happen because the columns are orthogonal vectors.

Thanks to PVAL and zhw for pointing corrections and for the useful comments.
