What are the values and types of the critical points, if any, of #f(x,y) = x(2-x-y)#?

I will assume by "critical points" you are referring to the values of #x# and #y# where the function is stationary.

First we need to find the partial derivatives of the function:

The #x# partial derivative:

#(partialf)/(partialx) = (2-x-y)+x(-1) = 2-x-y-x #
#=2-2x-y#

by use of the chain rule.

The #y# partial derivative:

#(partialf)/(partialy)=x(-1)=-x#

We now take both of these equations, and set them equal to 0 to obtain a pair of simultaneous equations which can then be solved to obtain one or more pairs of co-ordinates where our stationary point lies, so:

#2-2x-y=0#
#-x=0#

Immediately we can obviously see #x=0# from the bottom equation. That leaves us with:

#2-y=0#

for the top equation. This has the solution:

#y=2#

So we have only one stationary point at #(0,2)#. Here #f(0,2)=0#

In order to classify the point we now need to find the higher order derivatives, those will be:

#(partial^2f)/(partialx^2)=-2#
#(partial^2f)/(partialy^2)=0#
#(partial^2f)/(partialxpartialy)=-1#

We can classify the points by:

#(partial^2f)/(partialx^2)(partial^2f)/(partialy^2)-((partial^2f)/(partialxpartialy))^2 < 0# at #(x,y)#

then it will be a saddle point, or else if:

#(partial^2f)/(partialx^2)(partial^2f)/(partialy^2)-((partial^2f)/(partialxpartialy))^2 >0# as #(x,y)#

then it will be a maximum or minimum. Substituting the values of our derivative in we get:

#(2)(0)-(-1)^2=-1<0# So it is a saddle point. Hence our critical point is a saddle point at #(0,2)#.

I have included a contour plot of the function below (I did try a 3d plot at first but the shallowness of the gradient did not make the turning point clear enough). On the image the darker regions represent values of low #f(x,y)# and the brighter regions represent high #f(x,y)# in accordance with the plot legend.

We see that going from the bottom left to the top right you go up "over a hill" reaching a maximum at the calculated stationary point. Similarly going from the top left to the bottom right you go through the "bottom of a valley" with the minimum being at the calculated stationary point.

In other words, we notice that the contours from all directions converge at #(0,2)#, hence the stationary point of the function. Hopefully this helps to see the idea more clearly.