@ruhamdam In my answer, the second call to Basis was with the variables a[3],a[4],... substituted with x3, x4,... (without indices).

Your new worksheet also works for me in 1D; I did not try to find a formatting mistake in it, I have just converted it in 1D.
Please try the following:
Open a new worksheet with File > New > Worksheet Mode
and copy & paste the following:

with(Groebner);
with(PolynomialIdeals);
a := -9*w*v*t - 5/2*x*y*z*t + 6*w^3 + 5/2*x*z^2*w + 3*t^2*u + 3*y^3*t - 9*y*t - 4*y^2*z*w + 3/2*y*v*z^2 + 9*z*w - 1/2*z^3*u;
b := 2*x^4*z - 2*x^3*y^2 - 12*u^2*x*z + 36*u^2*y^2 - 48*u*v*x*y + 24*v^2*x^2 + 9*x^3 - 54*u^2;
c := 11*z^2*u*w + 20/3*x*z^2*y^2 - 12*t*y^2*v - 35*z*w^2*x + 24*y^2*z - 13/3*y^4*z - 90*w^2 - 11*u*y*z*t + 5*y*z*v*w + 10*x*v*t*z + 10*x*w*t*y - 7/3*x^3*x^2 + 22*w^2*y^2 + 3/2*t^2*x^2 - 15*x*z^2 - 3/2*v^2*z^2 + 36*t*v - 27*z;
d := 10*z*x*u*w + 10*y*u*v*z - 90*v^2 - 12*u*y^2*w + 20/3*z*x^2*y^2 + 11*x^2*v*t - 35*x*v^2*z + 22*v^2*y^2 + 3/2*u^2*z^2 - 7/3*x^3*z^2 - 3/2*x^2*w^2 + 36*u*w + 24*y^2*x - 13/3*y^4*x - 15*z*x^2 + 5*v*x*y*w - 11*x*y*u*t - 27*x;
e := 24*x*y*v*t - 87/2*x*v*z*w + 24*u*y*z*w + 33/2*x*u*z*t + 54*y^3 - 9*y^5 + 9/2*w*x^2*t + 51*y^2*v*w + 9/2*v*z^2*u - 7*x^2*z^2*y + 16*y^3*z*x - 45*u*y^2*t - 18*x*y*w^2 - 45*z*x*y - 18*y*v^2*z - 81*y + 81*u*t - 135*w*v;
f := 270*u^2*y*t - 3*x^2*y^2*w - 33/2*x^3*y*t + 57/2*x^3*z*w + 261*x*v^2*w - 72*u^2*z*w + 621*v*x*y + 54*x*u*w^2 + 405*z*x*u + 87/2*x^2*z^2*u - 219*y^3*v*x - 1134*u*y^2 + 270*u*y^4 + 108*x^2*w + 108*y*v^3 + 363/2*y*x^2*v*z - 171*x*u*v*t - 450*u*y*v*w - 285*u*y^2*z*x + 972*u;
g := 138*u*v*y*t - 129*x*y^2*v*z - 147*u*v*z*w + 187/2*x^2*y*z*w - 360*y^2*v + 62*v*y^4 + 27*x^2*t + 18*v^3*z - 48*x*u*w*t - 129/2*x*u*z^2*y - 243*u*y*z + 85*y^3*z*u - 21*x^2*y^2*t + 288*w*x*y + 288*z*x*v + 91/2*v*z^2*x^2 - 76*y^3*w*x - 24*x*v^2*t + 192*x*v*w^2 - 132*u*y*w^2 + 33*u^2*z*t + 9/2*x^3*z*t - 30*v^2*y*w + 486*v;
J := PolynomialIdeal(a, b, c, d, e, f, g);

then run it, pressing Enter. It should work.

@Mariusz Iwaniuk For such examples it is important to know the type of the solution. If we are looking for classical C^1 solutions, then indeed there are three of them. But if we want generalized solutions (continuous and differentiable except a finite number of points, or even absolutely continuous -- which both Maple and Matematica obtain sometimes) then there are infinitely many solutions. Is Mma able to find them? I don't think so.

@acer _C1 does not seem to be essential

restart;
eq1 := -ln(u)/2 + ln(3*u - 2)/6 - a - ln(x):
eq2 := -ln(u)/2 + ln(3*u - 2)/6 - a + ln(x):
solve(eq1, u) assuming x::real, a::real; # error
solve(eq2, u) assuming x::real, a::real; # ok, no RootOfs

Same if  ::real   is replaced with  > 0

@Carl Love Ok, but this ia actually a workaround for a bug.

Use simplify(%) assuming x>0;  after subs and you will have 0=0.

@student_1 You are maybe right, but there are many other plot options (e.g. linestyle) and I much prefer to see from Maple a focus on mathematical aspects (there are so many ...) rather than some time-consumming decorative details. 

@acer Yes, the solution should be

{-1 < x, x <= exp(2)-1},  {-1 - exp(2) < x, x < -1}

but it seems that it must be obtained (at least partially) by hand.

@ruhamdam 
 

D(y):=1: D(x):=x*y: D(z):=z^2: D(a):=0:
D(x*y + y*z + a*w);

      a*D(w) + x*y^2 + x + y*z^2 + z

@nm So, do you think that mathematically

LIMIT( diff(y(x), x) / y(x), y(x)=0) = 1   # ?

I have mentioned what odetest did. It is wrong.

[Edit] A math example.
Take y[n](x) = exp(-n*x), x in [1,2].
Then y[n] --> 0 (uniformly wrt x), but y[n]' / y[n] --> - oo.

@mmcdara SumOfDivisors is much older than 2015: numtheory:-sigma

@nm It is not Q which may fail (as you see, it is very simple). The problem is inherent for any CAS which cannot recognize sometimes a constant (or even 0) expression. Anyway, Q is much more reliable than solve. It is possible to use verify to check whether the result is constant or just evaluate it for some constants x,y (e.g. x=1, y=1 and then x=1,y=2) to obtain p; if p is the same, then check it directly; if not, then f is not isobaric.

@nm It works after a proper simplification:

Q((-(x*y)/2+sqrt(x^2*y^2-4*y)/2)*y, x,y):
expand(rationalize(%));
#                               -2

Edit. Q fails only for f(x,y) of the form f(x,y) = g(x)*y, 
but then g  must be g(x) = A/x, so, only the functions f(x,y) = A*y/x are not identified.

@nm The values given by P are correct too. 

restart;
f:=(x,y)->sqrt(x*y):
'p'=P(f(x,y), x,y):
p:=-1;
t^r=simplify(f(t*x, t^p*y)/f(x,y)) assuming t>0;
p:=2;
t^r=simplify(f(t*x, t^p*y)/f(x,y)) assuming t>0;
#                            p := -1
#                            t^r  = 1
#                            p := 2
#                            t^r  = t^(3/2)     

Actually, in this example p can be arbitrary!

@AHSAN I think you don't understand what I mean. When you try to compute e.g. int(x+y, x), you may obtain any of the expressions  x^2/2 + x*y,  x^2/2 + x*y + 7, x^2/2 + x*y + y^2 - ln(y),  x^2/2 + x*y + y^12 + sinh(y^2-2)  etc.
How are you going to interpret the plot, without knowing which expression is plotted?

 

@AHSAN I have told you that int(v,x) cannot be expressed by elementary functions. Maple probably tries to find a (huge and useless for you) expression containing sums over RootOfs (based on the residue theorem).
But even if B existed, it would be only modulo a function of y, so, your plot does not make much sense.