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My doubts here are related to the construction of 3D plots of Multivariable functions. As a consequence of that, I am going to keep these questions in this single post rather than creating several posts on the same topic.
It should be remarked that I have also read Mathematica's documentation and previous questions of Mathematica stack exchange. However, these latter did not help me.

With that said, suppose that we have the following multivariable function

$\displaystyle \psi(x,y)=\frac{-((2 (132 + 56 x^4 - 382 y + 394 y^2 - 171 y^3 + 26 y^4 + 6 x^3 (-47 + 31 y) + x^2 (608 - 782 y + 240 y^2) + x (-602 + 1130 y - 716 y^2 + 153 y^3)))}{((16 + 4 x^2 + 8 x (-2 + y) - 16 y + 5 y^2)^2 (5 x^2 + (3 - 2 y)^2 + x (-6 + 4 y))^2))}$

Based on the above, I ask the following questions:

  1. How may I construct a 3D plot of $\psi(x,y)$ so that positive values are represented by red colors and negative values are regarded to blue colors?

Here, I have tried to use the Plot3D command as follows,

Plot3D[-((2 (132 + 56 x^4 - 382 y + 394 y^2 - 171 y^3 + 26 y^4 + 
    6 x^3 (-47 + 31 y) + x^2 (608 - 782 y + 240 y^2) + 
    x (-602 + 1130 y - 716 y^2 + 153 y^3)))/((16 + 4 x^2 + 
    8 x (-2 + y) - 16 y + 5 y^2)^2 (5 x^2 + (3 - 2 y)^2 + 
    x (-6 + 4 y))^2)), {x, -1, 4}, {y, -1, 3}, ColorFunction -> Function[{x, y, z}, If[z > 0, Red, Blue]], AxesLabel -> (Style[#, Black, 20, Bold] & /@ {"x", "y", 
 "\[CapitalPsi]"}), TicksStyle -> Directive[Bold, Black],Epilog -> {Black, PointSize@Large, Point[{2, 1, 0}]}]

In this code, I have considered the ColorFunction command to represent the positive and negative values of $\psi(x,y)$. However, I am not quite sure whether this is indeed correct. As we may see, $\psi(x,y)$ is a negative function. Hence, should not we observe a predominantly blue color in the 3D plot of $\psi$?

  1. Is there a way to determine whether or not $\psi(x,y)$ is symmetric using Mathematica? That is to say, is the Swap command applicable to multivariable functions like $\psi$ ?

  2. How may one insert the point $(2,1)$ in the 3-D plot of $\psi(x,y)$? Here, I have applied the Epilog command, as one may see above. However, the point is not visiable.

Thanks and I look forward to hearing from you.

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    $\begingroup$ "How may I construct a 3D plot ... so that positive values are represented by red colors and negative values are regarded to blue colors?" - the trick I used here and in a few other answers is to use LogisticSigmoid[] + Blend[] in the ColorFunction; using a simpler example: Plot3D[Sin[x + Sin[y]], {x, -3, 3}, {y, -3, 3}, ColorFunction -> (Blend[{Blue, Gray, Red}, LogisticSigmoid[15 #3]] &), ColorFunctionScaling -> False] $\endgroup$ Jun 30, 2022 at 13:58

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  • Use MeshFunctions -> Function[{x, y, z}, z], Mesh -> {{0}}, MeshShading -> {Blue, Red}] to set positive values are represented by red colors and negative values are regarded to blue colors.

  • Use plot2 to draw meshs.

  • Use Graphics3D to draw a point in 3D.

SetOptions[Plot3D, PlotPoints -> 50, MaxRecursion -> 2, 
 AxesLabel -> (Style[#, Black, 15, Bold] & /@ {"x", "y", 
     "Ψ"}), TicksStyle -> Directive[Bold, Black]]; plot1 =
  Plot3D[-((2 (132 + 56 x^4 - 382 y + 394 y^2 - 171 y^3 + 26 y^4 + 
         6 x^3 (-47 + 31 y) + x^2 (608 - 782 y + 240 y^2) + 
         x (-602 + 1130 y - 716 y^2 + 153 y^3)))/((16 + 4 x^2 + 
          8 x (-2 + y) - 16 y + 5 y^2)^2 (5 x^2 + (3 - 2 y)^2 + 
          x (-6 + 4 y))^2)), {x, -1, 4}, {y, -1, 3}, 
  MeshFunctions -> Function[{x, y, z}, z], Mesh -> {{0}}, 
  MeshShading -> {Blue, Red}];
plot2 = Plot3D[-((2 (132 + 56 x^4 - 382 y + 394 y^2 - 171 y^3 + 
          26 y^4 + 6 x^3 (-47 + 31 y) + x^2 (608 - 782 y + 240 y^2) + 
          x (-602 + 1130 y - 716 y^2 + 153 y^3)))/((16 + 4 x^2 + 
           8 x (-2 + y) - 16 y + 5 y^2)^2 (5 x^2 + (3 - 2 y)^2 + 
           x (-6 + 4 y))^2)), {x, -1, 4}, {y, -1, 3}, 
   PlotStyle -> None, MeshStyle -> White];
plot3 = Graphics3D[{Yellow, PointSize@Large, Point[{2, 1, 0}]}];
Show[plot1, plot2, plot3]

enter image description here

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  • $\begingroup$ Hi @cvgmt, I hope you are doing well. Your answer is perfect! That's exactly what I was thinking. Thank you so much!!!! $\endgroup$
    – VH84
    Jun 30, 2022 at 14:42
  • $\begingroup$ @cvgmt.This year, I plan to write a scientific paper in which I will employ your suggestions. Based on that, I would like to acknowledge you. May you inform me your true real name and institution ? $\endgroup$
    – VH84
    Jun 30, 2022 at 14:55

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