I'm currently studying for a resit and I've been faced with this partial differentiation question:
If $z = f(y/x)$ show that $$x^2\frac{\partial^2 z}{\partial x^2}+2xy\frac{\partial^2 z}{\partial x\partial y} + y^2\frac{\partial^2 z}{\partial y^2} = 0$$
I can do the partial differentiation parts, but I am stuck from that point. It is late and I am tired so it is very possible I'm missing something glaringly obvious.
On
Factoring... $( x \frac{\partial}{\partial x} + y \frac{\partial}{\partial y} )^2 z = 0$. Now, just calculate the first order derivatives.
** EDIT **
More details: $z_x = f' (y/x) \cdot \frac{d}{dx} (y/x) = - (y/x^2) f'(y/x)$, and $z_y = f' (y/x) \cdot \frac{d}{dy} (y/x) = (1/x) f'(y/x)$, so $$ ( x \frac{\partial}{\partial x} + y \frac{\partial}{\partial y} )^2 z = (x \cdot - (y/x^2) f'(y/x) + y \cdot (1/x) f'(y/x))^2 = [f'(y/x) \cdot ( y/x - y/x ) ]^2 = 0. $$
We have $z=f(y/x)$. Thus,
$$\begin{align} x^2z_{xx}&=\,\,\,\,\,\,\,\,\left(\frac{y^2}{x^2}\right)f''(y/x)+\,\,\,\,\,\,\,\left(\frac{2y}{x}\right)f''(y/x) \tag 1\\\\ 2xyz_{xy}&=\left(-\frac{2y^2}{x^2}\right)f''(y/x)+\,\,\left(-\frac{2y}{x}\right)f''(y/x) \tag 2\\\\ y^2z_{yy}&=\,\,\,\,\,\left(\frac{y^2}{x^2}\right)f''(y/x)\tag 3 \end{align}$$
Adding $(1)-(3)$ yields the desired result!