Q:

$x^2 \frac{\partial^2{z} }{{\partial{x^2}}}+2xy \frac{\partial^2{z} }{{\partial{x}\partial{y}}}+y^2 \frac{\partial^2{z} }{{\partial{y^2}}}=0$

I want to solve this PDE. I seems like a classic problem but I don't know what to do when in second order PDE coefficients are polynomial themselves.
$$x^2 \frac{\partial^2{z} }{{\partial{x^2}}}+2xy \frac{\partial^2{z} }{{\partial{x}\partial{y}}}+y^2 \frac{\partial^2{z} }{{\partial{y^2}}}=0$$

A:

Let us translate the given PDE into polar coordinates.  Using $x = r \cos \theta$ and $y = r \sin \theta$, we get:
$$ \frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial r} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial r} = \cos \theta \frac{\partial z}{\partial x} + \sin \theta \frac{\partial z}{\partial y}.$$
Iterating this,
$$ \frac{\partial^2 z}{\partial r^2} = \cos \theta \left( \cos \theta \frac{\partial^2 z} {\partial x^2} + \sin \theta \frac{\partial^2 z} {\partial y \partial x}\right) + \sin \theta \left( \cos\theta \frac{\partial^2 z}{\partial x \partial y} + \sin\theta \frac{\partial^2 z}{\partial y^2}\right) = \\
\cos^2 \theta \frac{\partial^2 z}{\partial x^2} + 2 \cos \theta \sin \theta \frac{\partial^2 z}{\partial x \partial y} + \sin^2 \theta \frac{\partial^2 z}{\partial y^2} = \\
\frac{1}{r^2} \left( x^2 \frac{\partial^2 z}{\partial x^2} + 2xy \frac{\partial^2 z}{\partial x \partial y} + y^2 \frac{\partial^2 z}{\partial y^2} \right).$$
Therefore, the given PDE is equivalent to $r^2 \frac{\partial^2 z}{\partial r^2} = 0$ which has general solution of $z = f(\theta) + r g(\theta)$ (on any domain which is an annular sector not including the origin).

To relate this to the general solution from rafa11111's answer; or, to get a nice expression in terms of $x,y$: suppose our domain is contained either within the right half-plane or within the left half-plane.  Then $\theta = \tan^{-1}(y/x)$, so we have:
$$f(\theta) + r g(\theta) = f(\theta) + (r \cos \theta) (\sec \theta g(\theta)) = f(\tan^{-1}(y/x)) + x \left(\pm\sqrt{1 + (y/x)^2} g(\tan^{-1}(y/x)) \right)$$
(with the sign depending on the specific half-plane).
Therefore, if $F(m) = f(\tan^{-1}(m))$ and $G(m) = \pm \sqrt{1+m^2} g(\tan^{-1}(m))$ (with the chosen branch of $\tan^{-1}$ also depending on the half-plane) then $z = F(y/x) + x G(y/x)$.

As for how I came up with this approach: I first observed that the given PDE looks like an analogue of an Euler equation from ordinary differential equations.  This suggested that some solution of the form $z = x^\alpha y^\beta$ might work.  Plugging in this trial solution, we get:
$$\alpha (\alpha-1) x^\alpha y^\beta + 2 \alpha \beta x^\alpha y^\beta + \beta (\beta-1) x^\alpha y^\beta = [(\alpha + \beta)^2 - (\alpha + \beta)] x^\alpha y^\beta = 0.$$
Therefore, this does indeed give a solution whenever $\alpha + \beta \in \{ 0, 1 \}$.  Thus, we find particular solutions of the form $z = x^{-\beta} y^\beta = (y/x)^\beta$ and $z = x^{1-\beta} y^\beta = x (y/x)^\beta$.
Once we see this, it is reasonable to conjecture that in general, $f(y/x)$ and $x g(y/x)$ might be solutions - which is then straightforward to check.  Furthermore, by linearity of the equation, $z = f(y/x) + x g(y/x)$ is also a solution.  At this point, we heuristically have "enough degrees of freedom" that we think it could possibly be the general solution.  Also, from the appearance of $y/x$ in the solution, this suggests that the approach of converting the PDE into polar coordinates could be fruitful.  (Another approach would be to observe that the hypothesized general solution is linear on any ray from the origin, which would suggest examining the behavior of a solution on such curves $x = t \cos \alpha$, $y = t \sin \alpha$ for fixed $\alpha$.  The calculation for this approach would end up looking much the same.)

