How find this $\frac{yf_{y}-z}{f_{x}}+\frac{xf_{x}-z}{f_{y}}-xf_{x}-yf_{y}+x+y+z=C$ solution In plane $R^3$,Find $z=f(x,y)$, such the length of the portion of any tangent line to the astroid 
$$z=f(x,y)$$
cut off by the coordinate axes is constant $C$,
This problem is  from this post (when I answer it) How find a solution to this PDE $\frac{xf'_{x}}{f'_{y}}+\frac{yf'_{y}}{f'_{x}}+x+y=C$
my idea: I think this is anser is $$\sqrt{x}+\sqrt{y}+\sqrt{z}=C$$,because we easy to find this function is such it,
follow is my partial answer:
let $z=f(x,y)$,then the  tangent plane is 
$$f_{x}[X-x]+f_{y}[Y-y]=Z-z$$
so 
$$X=\dfrac{yf_{y}-z}{f_{x}}+x,Y=\dfrac{xf_{x}-z}{f_{y}}+y,Z=z-xf_{x}-yf_{y}$$
so
$$X+Y+Z=C\Longrightarrow \dfrac{yf_{y}-z}{f_{x}}+\dfrac{xf_{x}-z}{f_{y}}-xf_{x}-yf_{y}+x+y+z=C$$
then How find $z=f(x,y)?$ 
I know this PDE have one solution
$$\sqrt{x}+\sqrt{y}+\sqrt{z}=C $$ is such it,But Now How prove it?
Thank you,I fell this problem is interesting. I hope someone can help me,Thank you

my answer is :
firstly,we assume that:
$$X=\dfrac{yf_{y}-z}{f_{x}}+x=P+x,Y=\dfrac{xf_{x}-z}{f_{y}}+y=Q+y,Z=z-xf_{x}-yf_{y}=z-R$$
then,
$C^2=x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}=C+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}-P-Q+R$
$C(C-1)=(X+Y+Z)(X+Y+Z-1)=(X+Y+Z)^2-(X+Y+Z)=(x+y+z+P+Q-R)^2-(x+y+z+P+Q-R)=2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}-P-Q+R$$\Longrightarrow$$x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}=C^2$
$\Longrightarrow\sqrt{x}+\sqrt{y}+\sqrt{z}=x+y+z+P+Q-R$$\Longrightarrow$$\frac{f_{x}}{f_{y}}-{f_{x}}=\frac{1}{\sqrt{x}}-1$$,\frac{f_{y}}{f_{x}}-{f_{y}}=\frac{1}{\sqrt{y}}-1$$,\frac{{1}}{f_{x}}+\frac{1}{f_{y}}=1-\frac{1}{\sqrt{z}}$$\Longrightarrow$$f_{x}=-\frac{\sqrt{z}}{\sqrt{x}},f_{y}=-\frac{\sqrt{z}}{\sqrt{y}}$
hence, $C^2=C^2$ holds
