Let $f(x, y)=g(u, v)$ where the variables $\{x, y\}$ and $\{u, v\}$ are related by a smooth, invertible transformation. State the chain rule expressing the derivatives $\frac{\partial g}{\partial u}$ and $\frac{\partial g}{\partial v}$ in terms of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ and use this to deduce that

$$\frac{\partial^{2} g}{\partial u \partial v}=\frac{\partial x}{\partial u} \frac{\partial x}{\partial v} \frac{\partial^{2} f}{\partial x^{2}}+\left(\frac{\partial x}{\partial u} \frac{\partial y}{\partial v}+\frac{\partial x}{\partial v} \frac{\partial y}{\partial u}\right) \frac{\partial^{2} f}{\partial x \partial y}+\frac{\partial y}{\partial u} \frac{\partial y}{\partial v} \frac{\partial^{2} f}{\partial y^{2}}+H \frac{\partial f}{\partial x}+K \frac{\partial f}{\partial y}$$

where $H$ and $K$ are second-order partial derivatives, to be determined.

Using the transformation $x=u v$ and $y=u / v$ in the above identity, or otherwise, find the general solution of

$$x \frac{\partial^{2} f}{\partial x^{2}}-\frac{y^{2}}{x} \frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial f}{\partial x}-\frac{y}{x} \frac{\partial f}{\partial y}=0$$