We start by defining the Hamiltonian for this problem:

h[x_,y_]=x Cos[y]+y^2

Here's the default contour plot we get using ContourPlot:

ContourPlot[h[x,y],{x,-20,20},{y,-10,10},
ContourShading->False]

To see the level curves with saddle points, we'll need to carefully pick the level curve/contour values.  We have found equilbrium points including (π,π/2) and (-3π,3π/2).  Let's evaluate the Hamiltonian at each of these to find the corresponding level curve values

h[Pi,Pi/2]

h[-3Pi,3Pi/2]

These are multiples of (π^2)/4.  We'll use this fact in choosing level curve values in our contour plot.

There are two ways we can specify level curve values in the ContourPlot command.  Here's one way:

ContourPlot[h[x,y]==h[Pi,Pi/2],{x,-20,20},{y,-10,10},
ContourShading->False]

Same style but different level curve/contour value for the other equilibrium point we examined:

ContourPlot[h[x,y]==h[-3Pi,3Pi/2],{x,-20,20},{y,-10,10},
ContourShading->False]

This method is limited to one level curve/contour value.  To show more than one level curve/contour value, we can use the Contours option.  Here's an example:

ContourPlot[x Cos[y]+y^2,{x,-20,20},{y,-10,10},
Contours->{Pi^2/4,9*Pi^2/4,25Pi^2/4},
ContourShading->False]

Note that you can put the cursor on a level curve in any of these plots to see the corresponding level curve value.

We can use a Table command to generate a larger list of level curve/contour values.

ContourPlot[x Cos[y]+y^2,{x,-20,20},{y,-10,10},
Contours->Table[n Pi^2/4,{n,0,50}],
ContourShading->False]