The set of critical points is $\{(0,y):~y\in\mathbb{R}\}\cup\{(2,1)\}$. The Hessian of $f$ is $ Hf(x,y) = \begin{bmatrix} ye^{-x-y}(x^2-4x+2) & x(2-x)e^{-x-y}(1-y) \\ x(2-x)e^{-x-y}(1-y) & x^2e^{-x-y}(y-2) \end{bmatrix} $ The critical point $(2,1)$ is clearly a local maximum, since $Hf(2,1)$ has deteminant $8e^{-6}>0$ and trace $-6e^{-3}<0$.
For the other critical points the Hessian is of no use, since it is singular. Therefore we have to find a way around: first of all let us notice that $f(0,y)=0$ for all $y$. It is trivial to notice that \begin{align*} x^2ye^{-x-y}\geq 0 & \quad\text{for all $x$ and for all $y>0$} \\ x^2ye^{-x-y}\leq 0 & \quad\text{for all $x$ and for all $y<0$} \end{align*} This is sufficient to deduce that the points $\{(0,y):~y>0\}$ are all local minimums, $\{(0,y):~y<0\}$ are local maximums, and $(0,0)$ is a saddle.
[-0.1,0.1]\times[-0.1,0.1] to show the nature of $(0,0)$.">
If you wonder what a curve of critical points graphically means, think of a half pipe placed horizontally on the ground: the set of points that touch the ground is all made of local (global too...) minimums. For a more mathematical example, just consider the shape of the surface $f(x,y)=x^2$.