Eliminate variables $x$, $y$ and $z=(y^2-C)/x$ I am looking for a function of three variables $F(x,y,z)$ such that after the substitution
\begin{align*}
&x=x,\\
&y=y,\\
&z=\frac{y^2-C}{x}
\end{align*}
it turns out into a function of the constant $C$.
My idea was to derive a system of partial differential equations on $F$ and then solve them. By differentiate by $x,y$ the identity $F\left(x,y,\frac{y^2-C}{x}\right)=f(C)$ I have got 
\begin{cases}
\partial_x(F)+\partial_z(F) \cdot  \left(- \frac{y^2-C}{x^2} \right)=0,\\
\partial_y(F)+\partial_z(F) \cdot  \left( \frac{2y}{x} \right)=0.\\
\end{cases}
But Maple gives only a trivial solution $F(x,y,z)=const$   although one  solution is easy to guess 
$$
F(x,y,z)=xz-y^2.
$$
What is wrong with the system and how to find $F(x,y,z)$  without guessing?
 A: Using the method of characteristics, we can show that the PDE
$$
F_y + \tfrac{2y}{x} F_z = 0
$$
has solutions of the form
$$
F(x,y,z) = f(y^2 - xz) \, ,
$$
where $f$ is an arbitrary function.
Injecting this expression in the first PDE leads to
$$
\left(z - \tfrac{y^2-C}{x} \right) f'(y^2 - xz) = 0 \,
$$
which is always true due to the definition of $z$. Therefore, any function $F$ of the previous form will work, not only constant functions.
The proposed computation is actually useless. One notices that  $F(x,y,z)=f(C)$ with $C = y^2-zx$ gives the expected result (no differentiation and PDE solving needed).
A: Here are two ways of answering you question. The first method ignores the approach you were taking, but hopefully sets some ground rules and makes clear what the answer is.  The second addresses specifically your approach to make clear how it needs to be fixed up.
Method 1: 
You have said that that after your triple substitution, you want it to be the case that $F(x,y,z)$ will reduced to being $f(C)$, for some function $f$.  The function $f(C)$ could be completely arbitrary, or might be constrained in certain ways by personal choice (e.g. perhaps you want to restrict it to polynomials). Let's suppose $f$ is completely unconstrained so that, for example, $f(C)=\sin(\exp(C+4C^2))+1/C$ might be the form $f$ takes.  Given that rearrangement of your substitution shows that $C=y^2-xz$ we can see that in such an example we would have $$f(C)=\sin\left(\exp\left(y^2-xz+4\left(y^2-xz\right)^2\right)\right)+\frac 1{y^2-xz}$$ which is exactly the form that the corresponding $F$ would have to take:
$$F(x,y,z)=\sin\left(\exp\left(y^2-xz+4\left(y^2-xz\right)^2\right)\right)+\frac 1{y^2-xz}$$ in order to reach $f(C)$ after your substitution.
In other words, the functions $F(x,y,z)$ that meet your requirements are literally any functions you can build entirely out of $C$s, i.e. entirely out of $(y^2-xz)$ terms.   This is such a wide class of solutions ($f(C)$ can be as complicated as you like!) that they won't emerge as 'the' solution of a pair of simultaneous differential equations governed by (say) two constants of integration.  The answer is just far freer than that.  
Method 2:
Nonetheless, your desire seems to be to show that the functions $F$ which you seek are indeed able to be seen as solutions of a pair of differential equations, one of which expresses the lack of $x$ dependence in $f(C)$, while the other expresses the lack of $y$ dependence in the same.  It is certainly possible to proceed in that way, and since that is what you appear to want to do, let us do so:
I believe in showing you what went wrong in your approach, it will be helpful to change your notation slightly.  Specifically, part of your substitutions involves replacing $x$ with $x$ and $y$ with $y$.  Here you have used the same lower case letter to represent both the before- and the after-substitution variables. I would like to distinguish these carefully, so I will use $X$, $Y$ and $Z$ as the native (pre-subtitution) variables of $F$.  Since I am not using $z$ for a pre-substitution variable, so I will use it in pace of your $C$.  Your question re-stated in my notation becomes:
There is a function $$F(X,Y,Z)$$ which, after the substitutions $$X=x,\qquad (1a)\\Y=y,\qquad (1b)\\Z=\frac{y^2-z}{x}\qquad (1c)$$ must become a function $f(x,y,z)$ which will trivially equal its previous value (i.e. $F(X,Y,Z)=f(x,y,z)$) but, critically, must also (by choice) satisfy two extra constraints: $$\left.\frac{\partial  f}{\partial  x}\right|_{y,z}=0, \qquad\text{and}\qquad\left.\frac{\partial  f}{\partial  y}\right|_{x,z}=0\qquad(2)$$ which ensure that $f(x,y,z)$ is just a function of $z$.
By the chain rule for partial derivatives, we have:
$$
\left.\frac{\partial  f}{\partial  x}\right|_{y,z} 
= 
\left.\frac{\partial  F}{\partial  X}\right|_{Y,Z} 
\left.\frac{\partial  X}{\partial  x}\right|_{y,z} 
+
\left.\frac{\partial  F}{\partial  Y}\right|_{X,Z} 
\left.\frac{\partial  Y}{\partial  x}\right|_{y,z} 
+
\left.\frac{\partial  F}{\partial  Z}\right|_{X,Y} 
\left.\frac{\partial  Z}{\partial  x}\right|_{y,z} 
\qquad(3)
$$
and
$$
\left.\frac{\partial  f}{\partial  y}\right|_{x,z} 
= 
\left.\frac{\partial  F}{\partial  X}\right|_{Y,Z} 
\left.\frac{\partial  X}{\partial  y}\right|_{x,z} 
+
\left.\frac{\partial  F}{\partial  Y}\right|_{X,Z} 
\left.\frac{\partial  Y}{\partial  x}\right|_{y,z} 
+
\left.\frac{\partial  F}{\partial  Z}\right|_{X,Y} 
\left.\frac{\partial  Z}{\partial  y}\right|_{x,z} 
\qquad(4)
$$
and so using our substitutions (1) and our constraints (2) we find that our chain rules (3) and (4) reduce to:
\begin{align}
0
&=
\left.\frac{\partial  F}{\partial  X}\right|_{Y,Z}
.1
+
\left.\frac{\partial  F}{\partial  Y}\right|_{X,Z}
.0
+
\left.\frac{\partial  F}{\partial  Z}\right|_{X,Y}
.
\left( - \frac{y^2-z}{x^2} \right), \qquad\text{and} \qquad(5)
\\
0
&=
\left.\frac{\partial  F}{\partial  X}\right|_{Y,Z}
.0
+
\left.\frac{\partial  F}{\partial  Y}\right|_{X,Z}
.1
+
\left.\frac{\partial  F}{\partial  Z}\right|_{X,Y}
.
\left( \frac{2 y} {x} \right)\qquad\qquad(6)
\end{align}
or equivalently
\begin{align}
0
&=
\left.\frac{\partial  F}{\partial  X}\right|_{Y,Z}
-
\left.\frac{\partial  F}{\partial  Z}\right|_{X,Y}
  \frac{y^2-z}{x^2}, \qquad\text{and} \qquad(7)
\\
0
&=
\left.\frac{\partial  F}{\partial  Y}\right|_{X,Z}
+
2 \left.\frac{\partial  F}{\partial  Z}\right|_{X,Y}
 \frac{ y} {x} \qquad\qquad(8)
\end{align}
or equivalently
\begin{align}
0 
&=
\left.\frac{\partial  F}{\partial  X}\right|_{Y,Z}
-
\left.\frac{\partial  F}{\partial  Z}\right|_{X,Y}
\frac Z X,\qquad\text{and}\qquad(9)
\\
0 
&=
\left.\frac{\partial  F}{\partial  Y}\right|_{X,Z}
+
2
\left.\frac{\partial  F}{\partial  Z}\right|_{X,Y}
\frac Y X.\qquad\qquad(10)
\end{align}
Note now carefully that although (8) looks a bit like (10) ... (indeed it would look identical had we not labelled $X$ and $x$ differently, etc) ... the same is not true for (7) and (9).  Equation (7) doesn't look much like (9) at all!  That is to say, if we replaced $(x,y,z)\rightarrow(X,Y,Z)$ in (7) we do not get (9) -- nor should we -- instead we (would) get:
$$
0
=
\left.\frac{\partial  F}{\partial  X}\right|_{Y,Z}
-
\left.\frac{\partial  F}{\partial  Z}\right|_{X,Y}
  \frac{Y^2-Z}{X^2}, \qquad\text{and} \qquad(\text{broken version of 7})
$$
From my understanding of the question as originally posted, the questioner was (incorrectly) asking Maple to solve (broken version of 7) simultaneously with (10). What should be solved is (9) simultaneously with (10), since in (9) and (10) all the variables are of the same type -- demonstrated by their consistent capitalization.
We can check that the above differential equation is sensible by, for example, setting $F(X,Y,Z)$ equaul to $XZ-Y^2$ ... one of the questioner's guessed solutions.  With $F(X,Y,Z)=XZ-Y^2$ we have 
$\left.\frac{\partial  F}{\partial  X}\right|_{Y,Z}=Z$, 
$\left.\frac{\partial  F}{\partial  Y}\right|_{X,Z}=-2 Y$ and
$\left.\frac{\partial  F}{\partial  Z}\right|_{X,Y}=X$, and so see that the right hand sides of (9) and (10) are trivially satisifed.  The same is not true if you take the original questioner's differential equations:
\begin{cases} \partial_x(F)+\partial_z(F) \cdot \left(- \frac{y^2-C}{x^2} \right)=0,\qquad(15)\\ \partial_y(F)+\partial_z(F) \cdot \left( \frac{2y}{x} \right)=0.\\ \end{cases} and substitute in his/her guess $$ F(x,y,z)=xz-y^2 $$ since (in the original questioner's notation) we have $\partial_x (F)=z$ and $\partial_z(F)=x$ and so the left hand side of (15) becomes $$z+x\left(-\frac{y^2-C}{x^2}\right)$$ which is not zero, and has no reason to be zero.
In summary, the moral of the story is that one should be careful not to confuse the variables pre- and post-substitution if one wants to proceed using an approach more like method 2 than like method 1.
