Okay, this question might seem odd and the theory might not work due to the definition of a derivative, but I keep noticing something of when I am taking the second derivative of a function where there needs to be change of variable. Observe, for some function $u(x,t)$. Find the second derivative of u with respect to $\gamma$ and then $\alpha$ or vice versa. Let $\gamma=x-ct, \alpha=x+ct$.
If I have $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \gamma}+\frac{\partial u}{\partial \alpha}$, then take the second derivative of that with respect to $x$, I get $\frac{\partial ^2 u}{\partial x^2}=\frac{\partial}{\partial x}(\frac{\partial u}{\partial \gamma}+\frac{\partial u}{\partial \alpha})=\frac{\partial ^2 u}{\partial \gamma^2}+\frac{\partial ^2 u}{\partial \alpha^2}+2\frac{\partial ^2 u}{\partial \gamma \partial \alpha}$.
There are two ways to do this problem I noticed.
$\frac{\partial ^2 u}{\partial x^2}=\frac{\partial}{\partial x}(\frac{\partial u}{\partial \gamma}+\frac{\partial u}{\partial \alpha})=\frac{\partial}{\partial \gamma}(\frac{\partial u}{\partial \gamma}+\frac{\partial u}{\partial \alpha})+\frac{\partial}{\partial \alpha}(\frac{\partial u}{\partial \gamma}+\frac{\partial u}{\partial \alpha})=\frac{\partial ^2 u}{\partial \gamma^2}+\frac{\partial ^2 u}{\partial \alpha^2}+2\frac{\partial ^2 u}{\partial \gamma \partial \alpha}$.
($\frac{\partial u}{\partial x})^2=(\frac{\partial u}{\partial \gamma}+\frac{\partial u}{\partial \alpha})^2=(\frac{\partial u}{\partial \gamma}+\frac{\partial u}{\partial \alpha})*(\frac{\partial u}{\partial \gamma}+\frac{\partial u}{\partial \alpha})=\frac{\partial ^2 u}{\partial \gamma^2}+\frac{\partial ^2 u}{\partial \alpha^2}+2\frac{\partial ^2 u}{\partial \gamma \partial \alpha}$.
Is this even possible? I understand why 1 is possible because of the change in variable with the chain rule and mainly the derivatives of $\gamma, \alpha$ with respect to $x$ are 1. I think it may just be a special case since the derivatives of $\gamma, \alpha$ with respect to $x$ are simply 1. No other case would do this.
For more information on the overall problem, please see $\frac{\partial^2 u}{\partial t^2}-c^2\frac{\partial^2u}{\partial x^2}=0\text{ implies } \frac{\partial^2 u}{\partial z \partial y}=0$