Solution of partial differential equation Solve the differential equation,
$$
z=\frac{\partial z}{\partial x}x + \frac{\partial z}{\partial y}y+ (\frac{\partial z}{\partial x})^2 + \frac{\partial z}{\partial x}\frac{\partial z}{\partial y}+  (\frac{\partial z}{\partial y})^2.$$
Can you please solve and explain the solution assuming that I don't have any idea.
 A: For any function $\phi$ and coordinate variable $w$, we will use
$\partial_w$ and $\phi_w$ as a shorthand of the differential operator $\frac{\partial}{\partial w}$ and the partial derivative $\partial_w \phi = \frac{\partial\phi}{\partial w}$.
Consider following change of variables
$$(x,y) = \left(\frac{\sqrt{3}u+v}{2}, \frac{\sqrt{3}u-v}{2}\right)\quad\iff\quad
(u,v) = \left(\frac{x+y}{\sqrt{3}},x-y\right)$$
We have $$
\begin{cases}
\partial_u &= \frac{\sqrt{3}}{2}(\partial_x + \partial_y)\\
\partial_v &= \frac12(\partial_x - \partial_y)
\end{cases}
\;\;\implies\;\;
\begin{cases}
u z_u + v z_v &= x z_x + y z_y\\
z_u^2 + z_v^2 &= \frac34(z_x + z_y)^2 + \frac14(z_x - z_y)^2 = z_x^2 + z_xz_y + z_y^2
\end{cases}
$$
Let $\rho^2 = u^2 + v^2$ and $\varphi = z + \frac{\rho^2}{4}$. The PDE at hand
can be transformed to
$$\begin{array}{rrcl}
     & z &=& uz_u + v z_v + z_u^2 + z_v^2\\
\iff & z + \frac{\rho^2}{4} &=& \left(z_u + \frac{u}{2}\right)^2 + \left(z_v + \frac{v}{2}\right)^2\\
\iff & \varphi &=& \varphi_u^2 + \varphi_v^2\tag{*1}
\end{array}
$$
The last equation in $\varphi$ has a trivial solution $\varphi \equiv 0$. This leads to one solution for $z$:
$$z = -\frac{\rho^2}{4} = -\frac{u^2+v^2}{4} = -\frac{x^2 - xy + y^2}{3}\tag{S.1}$$
On those places where $\varphi \ne 0$, we have $\varphi > 0$. Introduce another function $\psi$ such that $$\psi = \sqrt{4\varphi}\quad\iff\quad \varphi = \frac{\psi^2}{4}$$
the function $\psi$ will satisfy
$$|\nabla \psi|^2 = \psi_u^2 + \psi_v^2 = 1\tag{*2}$$
Conversely, given any solution of $(*2)$, the PDE in $z$ will have a solution of the from 
$$z = \frac14(\psi^2 - \rho^2)\tag{*3}$$
For example, 


*

*Set $\psi$ to the distance between $(u,v)$ and some 
fixed point $(-\sqrt{3}(a+b),-(a-b))$, one obtain following solution found by JJacquelin:


$$\begin{align}
z &= \frac14\left((u+\sqrt{3}(a+b))^2 + (v+(a-b))^2 - u^2 - v^2\right)\\
  &= a x + b y + a^2 + ab +b^2
\end{align}\tag{S.2}
$$


*

*Set $\psi$ to any affine function with slope $1$ along direction $\theta$ in $uv$-plane, one find a family of quadratic surfaces as solutions of $z$:
$$\begin{align}
z &= \frac14\left[ \left(\cos\theta u + \sin\theta v + m\right)^2 - (u^2+v^2)\right]\\
 &= -\frac14\left[ \left(-\sin\theta u + \cos\theta v\right)^2 - 2m \left(\cos\theta u + \sin\theta v\right) - m^2 \right]\tag{S.3}
\end{align}
$$
where $\theta, m$ are constants.


These $3$ solutions $(S.1)$, $(S.2)$ and $(S.3)$ are all the global solutions I can find. If one drop the requirement that $z$ is defined for all $(x,y)$, there are other ways to construct more solutions. 
Let


*

*$K$ be any convex body in the plane bounded by some smooth curve.

*Let $d_K(p)$  be the distance of any point $p$ to $K$. i.e.
$$d_K(p) = \inf\{ | p - q | : q \in K \}$$


Outside $K$, $d_K(p)$ is smooth, positive and $|\nabla d_K| = 1$.
For any constant $\psi_0 \ge 0$, we can set $\psi$ to $d_k + \psi_0$ and
use $(*3)$ to construct a solution for the PDE over there.
A: HINT (partial answer) :
Search for a particular linear solution on the form $z=ax+by+c$
Puting into the PDE leads to 
$$z=ax+by+a^2+ab+b^2 \quad\text{any }a\:,\: b.$$
