Solution of pde $\frac{\partial z}{\partial x}\cdot \frac{\partial z}{\partial y}=1$ with $z(0,0)=0$ The problem is to solve $\frac{\partial z}{\partial x}\cdot \frac{\partial z}{\partial y}=1$  with $z(0,0)=0$ and find $z(0,1)$.
I used Charpit's method to get $z(x,y)=kx + \frac{y}{k}+c$ as its solution. Now the initial condition gives $c=0$. This gives $z(x,y)=kx + \frac{y}{k}$ but the arbitrary constant is still present even after incorporating the initial condition and this is my point of confusion. Then $z(0,1)=\frac{1}{k}\ne 0$ if $k$ is non-zero.
Added 1- I realized that the solution can also take the form $z=ky+\frac{x}{k}$ which gives me $z(0,1)=k$.

What is correct? $z(0,1)=\frac{1}{k}\ne 0$ or $z(0,1)=k$?

Am I correct? Any hints where I am wrong?
Edit- After few searches I found a way to determine the arbitrary constant 'k' which proceeds as follows:
Solve the $k$-quadratic $k^2x-kz+y=0$ for $k$ and substitute this $k$ in $z=k x+\frac{y}{k}$ which is now free from $k$.
I suspect this approach as $k$ thus obtained is no more a constant and the reason for determining $k$ in this way is not justified. I don't know it is correct or not. Hoping that someone answer my doubt!
 A: There is another class of solutions to this PDE.
We try separation-of-variables and guess a solution of the form $z(x,y) = f(x) g(y) + c$:
\begin{align*}
\frac{\partial z}{\partial x} \cdot \frac{\partial z}{\partial y}
&\;=\; f'(x) g(y)\cdot f(x) g'(y)\\
&\;=\; f'(x) f(x)\cdot g'(y) g(y) \;=\; 1
\end{align*}
The last expression above can only be true for all $x$ and $y$ if the terms in $x$ are equal to some constant $k$ and the terms in $y$ are equal to $1/k$:
\begin{align*}
f'(x) f(x) &\;=\; k \;\;\;\quad\rightarrow\quad f(x) \;=\; \sqrt{2(k x \,+\, a)}\\[0.1in]
g'(y) g(y) &\;=\; 1/k \quad\rightarrow\quad g(y) \;=\; \sqrt{2(y/k \,+\, b)}
\end{align*}
Here $a$ and $b$ are unknown constants. The solution to the PDE is thus:
\begin{equation*}
z(x,y) \;=\; f(x) g(y) + c \;=\;
2\sqrt{(k x \,+\, a)(y/k \,+\, b)} \,+\, c
\end{equation*}
The boundary condition $z(0,0) = 0$ implies that the unknown constants $a$, $b$, and $c$ must obey $2\sqrt{ab} + c = 0$.
A: $$\frac{\partial z}{\partial x}\cdot \frac{\partial z}{\partial y}=1\quad\text{ with }
\quad  z(0,0)=0 \tag 1$$
$$\frac{\partial^2 z}{\partial x^2}\cdot \frac{\partial z}{\partial y}+\frac{\partial z}{\partial x}\cdot \frac{\partial^2 z}{\partial x\partial y}=0$$
Change of function : $\quad u(x,y)=\frac{\partial z}{\partial x}$
$$\frac{\partial u}{\partial x}\left(-\frac{1}{u} \right)+u\frac{\partial u}{\partial y}=0$$
$$\frac{\partial u}{\partial x}-u^2\frac{\partial u}{\partial y}=0 \tag 2$$
system of characteristic ODEs : $\frac{dx}{1}=\frac{dy}{-u^2}=\frac{du}{0}$
First characteristic equation from $\frac{du}{0}=$finite function :
$$u=c_1$$
Second characteristic equation from $\frac{dx}{1}=\frac{dy}{-c_1^2}\quad;\quad x+\frac{y}{c_1^2}=c_2$
$$x+\frac{y}{u^2}=c_2$$
General solution of Eq.$(2)$ on the form of implicite equation $c_2=F(c_1)$ :
$$x+\frac{y}{u^2}=F(u)$$
$F$ is an arbitrary function to be determined by some boundary condition.
CONDITION :
The only condition specified in the wording of the question is $z(0,0)=0$. Generally  the boundary condition is given on a curve. On a point only  is not sufficient to a well posed condition for Eq.$(2)$. Thus we expect an infinity of solutions.
EXAMPLE :
Arbitrary choosing the function $F(X)=a+\frac{b}{X^2}$
$x+\frac{y}{u^2}=a-\frac{b}{u^2}$
$u=\sqrt{\frac{-b-y}{x-a}}$ 
$z(x,y)=\int\sqrt{\frac{-b-y}{x-a}}dx=2\:\sqrt{(-b-y)(x-a)}+c$
This function is a solution of the PDE $(1)$ . We determine $c$ in order to satisfy the condition $z(0,0)=0$
$$z(x,y)=2\:\sqrt{(b+y)(a-x)}-2\:\sqrt{ab}$$
In particular $\quad z(0,1)=2\:\sqrt{b(a-1)}-2\:\sqrt{ab}$
Of course this is only an example. Other choice of the function $F(X)$ would lead to other results. Note that many would be not explicit and/or complicated depending on the kind of function $F(X)$.
In conclusion don't be surprized to not find a unique result for $z(0,1)$. Probably this is due to a typo in the wording of the problem. Especially a boundary condition such as $z(0,0)=0$ seems suspect.
