@nm This is obviously a bug. Actually a regression bug, inexistent in Maple 2023 (and an elementary one, the expression being continuous w.r.t. x).
Unfortunately, today is much more difficult to find an explanation than 20 (or even 10)  years ago! 

@nm  OK, then I suggest to start some other similar questions such as about:

e:=-tanh(sqrt(2)*(a*x+b)):
limit(e,x=0): expand(%);

and

e:=-tanh(sqrt(2)*(a*x+b)):
limit(e,x=0): convert(%, exp): %;

etc.

I did not delete it, but why don't you continue your previous question having exactly the same limit?
limit not evaluating - MaplePrimes

@C_R  Actually  abc?:= something;   or better  abc? := something;
is an assignment to the variable abc?
Note that ? is a legit character in a symbol (but not as a first character).

The behavior is not new for me. Unfortunately it appeared a few years ago.

@Kitonum  Strange that evala works. Probably just by accident.

Equation or do you mean a function?

If it's a function, do you want the expansion in a series in all the variables (provided it exists)? Around (0,0,...,0)?

@annarita Why don't you try?

@annarita Yes, it is possible, if f is defined:

f := (x, y) -> piecewise(x=0 and y=0,  0, (x^3*y - x*y^3)/(y^2 + x^2)):

(or better, use And(x=0, y=0)  instead of x=0 and y=0).

Then:

D[1](f)(0,0);
D[2](f)(0,0);

are both 0.

I have appended the answer in your previous question. 
You should delete this one.

@salim-barzani 

Remarks.

1. Your new function phi may generate errors (divisions by 0).
It must be rewritten or called inside try - catch - end try.

2. It seems you just want to learn programming. In this case don't choose such a large output.
ou have increased the numbers of loops to 8, so a potential  4^8 = 65536 rows in the answer. Difficult to analyse.

3. For such a large output the (poor!)  Java interface is very slow and may even crash.
It is recommended to generate an Array and place the rows there.

restart;
phi := (p__1, p__2, p__3, p__4, q__1, q__2, q__3, q__4, xi) -> (p__1*exp(q__1*xi) - p__2*exp(q__2*xi))/(p__3*exp(q__3*xi) + p__4*exp(q__4*xi)):
xi := x:   # it should be numeric for your evalf!

A:=Array([]):
nr:=0:
for p__1 in [1, -1, I, -I] do
for p__2 in [1, -1, I, -I] do 
for p__3 in [1, -1, I, -I] do 
for p__4 in [1, -1, I, -I] do 
for q__1 in [1, -1, I, -I] do 
for q__2 in [1, -1, I, -I] do 
for q__3 in [1, -1, I, -I] do 
for q__4 in [1, -1, I, -I] do 
  try
    result1 := evalf(phi(p__1, p__2, p__3, p__4, q__1, q__2, q__3, q__4, xi));
    catch "numeric":
    result1:=0;
  end try;
  if result1=0 then next else
    A(++nr):= [arg=[p__1, p__2, p__3, p__4, q__1, q__2, q__3, q__4, xi], ans=result1] fi; 
end do; end do; end do; end do; end do; end do; end do; end do:

A[1..2];

upperbound(A);
                             57600
 

@mmcdara  Something seems to be  wrong in your example.

restart;
eq:=(a*x+b*y)*diff(u(x, y), x) + (c*x+d*y)*diff(u(x, y), y):
#  "For this problem to have a solution you must have a=-d"  ??
EQ:=eval(eq, [a=4,b=2,c=1,d=3]):
SOL:=pdsolve(EQ): lprint(%);
# u(x,y) = f__1((x^2+2*x*y+y^2)/(2*y-x)^5)

eval(EQ,SOL): simplify(%);
#            0

restart;
ex := sqrt( RootOf((1 - sqrt(_Z^3 + _Z^2 - _Z + 1))) ):
radnormal(ex);
# Error, (in RootOf) _Z occurs but is not the dependent variable

@JAMET Maybe this will give you an idea for future posts. Your undocumented and unexplained problem received an undocumented and unexplained answer. It seems that you do not like it (even if it should be easy to fix).

@janhardo Not necessarily a=a. The expressions are evaluated [+ automatic simplifications], but not simplified.

restart;
a:=x+1; b:=a+8; c:=a-9+b-x;
evalb(a=c);    # true