If an expression has more than one variable, then we may meed more than one quantifier( we could also eliminate variable via assignment)
In this case,
$\exist x\exist y(x^2+y^2=25)$ is really $\exist x\underbrace{\Big[\exist y(x^2+y^2=25)\Big]}_{x\text{ is not a variable anymore}}$,
We verify $\exist$ with an example.
In this case, for $x=4$
This is an example provide $\exist y(4^2+y^2=25)$is true which it is since $3$ is an example.
If all the quantifiers are the same ($\text{i.e. all }\forall\text{ or all }\exist$) then, the order doesn’t actually matters.
One example pair $(x,y)$ can verify $\exist x\Big[\exist yP(x,y)\Big]\text{ or }\exist y\big[\exist x P(,y)\big],$ similarly for $\forall$.
If there is a mixture, order can matter.
Ex. $\forall x \exist y(x+y=0)$ universal is $\R$, This is supposed to be $\forall x\Big[\exist y(x+y=0)\Big]\ne0$
The whole compound expression is true because the inner expression is true no matter what x we try to turn into a counterexample.
OTOH
$\exist y\forall x(x+y=0)$ is false
here $x=1-y$ fails $x+y= (1-y)+y =1\ne 0$
What ever… example…inner
OTOOH,
There is nothing inherently wrong with $\exist y \forall x$