@Markiyan Hirnyk  The code works under these refinements.

@Markiyan Hirnyk  

simplify(diff(u, y), {diff(u, x) = 0});

                                 0

You wrote  "another root  that i known, are : 0.165237712988657e-1    and     .103583272213766    and    .290071279318035"

See

plot(g, x = 0 .. .4);

@vv  You wrote  "In the "exterior" of this hyperbola, the solution is on a curve (also a hyperbola)". But 

{x=(-4*y-6)/(y+2), y=y}  is the same hyperbola. 

@nm  When the  unapply  command is called, there is an immediate calculation, and if  ->  command is called, then the calculation is delayed.

Here is an example:

f:=unapply(int(x^2, x=0..a),a);

g:=a->int(x^2, x=0..a);

f(2);

g(2);

Addition: Here an example where the first command works, and the second command does not work: 

f:=unapply(rsolve({u(n+1)=u(n)+2, u(1)=1}, u(n)), n);

g:=n->rsolve({u(n+1)=u(n)+2, u(1)=1}, u(n));

f(3);

g(3);

The reason is obvious - the premature calculation.

@Bendesarts  Here is the correction my previous plotting (I did not notice that the lines at the zeros of the function to be built only in the points of increasing of the function) . I also added vertical lines at points of extreme.

restart;

v := t->H*sin(w*t):

L := 0.080: H := 0.020: Vf := 0.3: w := 10:

zeros1 := sort([RootFinding[Analytic](v(t), t, re = .5 .. 2, im = -1 .. 1)]):
zeros10 := select(t->D(v)(t)>0, zeros1);

zeros2 := sort([RootFinding[Analytic]((D(v))(t), t, re = .5 .. 2, im = -1 .. 1)]):
zeros20 := selectremove(t->D[1, 1](v)(t) < 0, zeros2);

plot([v(t), seq([zeros10[k], t, t = -0.01 .. 0.01], k = 1 .. 3), seq([zeros20[1, k], y, y = v(zeros20[1, k])-0.01 .. v(zeros20[1, k])+0.01], k = 1 .. 2), seq([zeros20[2, k], y, y = v(zeros20[2, k])-0.01 .. v(zeros20[2, k])+0.01], k = 1 .. 2)], t = 0.5 .. 2, color = [black, yellow$3, red$2, blue$2], linestyle = [1, 3$7], thickness = [1, 2$7]);

@Rouben Rostamian   Examples in the help and my own experiments with  Basis  command confirmed it (remember that the initial rows are assumed to be linearly independent). Of course if you need to be sure absolutely, it's necessary to study the source code of  Basis  procedure.

Addition: of course, the addition of linearly independent unit vectors to the initial rows can be done without  Basis command, if first we apply  LinearAlgebra[GaussianElimination]  or  LinearAlgebra[ReducedRowEchelonForm]  commands.

@shzan  I am busy now. My answer to your question later today.

@shzan   First apply  LinearAlgebra[GaussianElimination]  command.

@H-R  Replace  2*Pi  to  6.28  in the code .  I do not know the cause of the error

@vv  for this workaround. But the error  (with  `if`)  remains. I would like to in the future releases the situation has been corrected.

@vv   I tried to apply  nth-term test for divergence   for your function f :

f:=unapply(`if`(n<10^10, 0, 10^10), n):

limit(f(n), n=infinity);

    Error, (in limit/dosubs) invalid input: if expects 3 arguments, but received 1

@Markiyan Hirnyk 

For better perception must reduce the number of contours and the range for  y :

ContoursWithLabels(2*(2-x)^2*exp(-x^2-(y+1)^2)-(15*((1/5)*x-x^3-y^3))*exp(-x^2-y^2)-(1/3)*exp(-(x+1)^2-y^2), -2 .. 2, -2.5 .. 2.5, {seq(-6 .. 6, 1)}, [color = black, axes = box], Coloring = [colorstyle = HUE, colorscheme = ["Cyan", "Red"], style = surface]);

 @H-R

My answer see in this  post

@farhad-b 

1) You wrote "is there any different between f:= x(t)^2 + y(t)^2:  and  f:=t->x(t)^2+y(t)^2" ?   About that distinction, I wrote in detail above.

2) I have never worked in Matlab and do not export to Matlab, so I can not give any advice.